Class 12 Mathematics - NCERT Mathematics Part II
NCERT Solutions for Class 12 Maths Chapter 7: Integrals
Chapter 7 is the largest and most important chapter in Class 12 Maths. It covers indefinite integrals (anti-derivatives), integration methods (substitution, partial fractions, by parts), and definite integrals (fundamental theorem of calculus, properties). This chapter is part of the Calculus unit which carries 35 marks in the CBSE Class 12 board exam 2027. Integrals alone typically contribute 15-18 marks making it the highest-scoring single chapter.
Syllabus Update 2026-27
As per the CBSE 2026-27 syllabus, Chapter 7 has 11 exercises plus a Miscellaneous Exercise. Ex 7.1 (22 Qs) basic integration, Ex 7.2 (39 Qs) substitution, Ex 7.3 (24 Qs) trig identities, Ex 7.4 (25 Qs) special functions, Ex 7.5 (23 Qs) partial fractions, Ex 7.6 (24 Qs) by parts, Ex 7.7 (11 Qs) special types, Ex 7.8 (6 Qs) limit of sum, Ex 7.9 (22 Qs) fundamental theorem, Ex 7.10 (10 Qs) definite by substitution, Ex 7.11 (21 Qs) properties, Misc (44 Qs). Total ~271 questions.
Table of Contents
7.1Exercise 7.1 - Basic Integration (Anti-derivatives)
10 key questions shown below covering all standard integral formulas.
Q1Find an anti-derivative (or integral) of the following functions by the method of inspection: (i) sin 2x (ii) cos 3x (iii) e^(2x) (iv) (ax + b)² (v) sin 2x - 4e^(3x)
By inspection, we find the function whose derivative is the given function.
(i) sin 2x:
d/dx(-cos 2x) = 2sin 2x.
So anti-derivative of sin 2x = (−1/2)cos 2x + C
(ii) cos 3x:
d/dx(sin 3x) = 3cos 3x.
So anti-derivative of cos 3x = (1/3)sin 3x + C
(iii) e^(2x):
d/dx(e^(2x)) = 2e^(2x).
So anti-derivative of e^(2x) = (1/2)e^(2x) + C
(iv) (ax + b)²:
d/dx[(ax+b)³] = 3a(ax+b)².
So anti-derivative of (ax+b)² = (ax+b)³/(3a) + C
(v) sin 2x − 4e^(3x):
Anti-derivative = (−1/2)cos 2x − 4 × (1/3)e^(3x) + C = (−1/2)cos 2x − (4/3)e^(3x) + C
Q2Find the following integrals: (i) ∫(4e^(3x) + 1) dx (ii) ∫x²(1 − 1/x²) dx (iii) ∫(ax² + bx + c) dx
(i) ∫(4e^(3x) + 1) dx:
= 4 × e^(3x)/3 + x + C
= (4/3)e^(3x) + x + C
(ii) ∫x²(1 − 1/x²) dx = ∫(x² − 1) dx:
= x³/3 − x + C
= x³/3 − x + C
(iii) ∫(ax² + bx + c) dx:
= a(x³/3) + b(x²/2) + cx + C
= ax³/3 + bx²/2 + cx + C
Q5Find the following integrals: (i) ∫(x³ − x² + x − 1)/(x − 1) dx (ii) ∫(1 + x)√x dx
(i) ∫(x³ − x² + x − 1)/(x − 1) dx:
Factor: x³ − x² + x − 1 = x²(x − 1) + 1(x − 1) = (x − 1)(x² + 1).
So (x³ − x² + x − 1)/(x − 1) = x² + 1.
∫(x² + 1) dx = x³/3 + x + C
(ii) ∫(1 + x)√x dx = ∫(√x + x√x) dx = ∫(x^(1/2) + x^(3/2)) dx:
= x^(3/2)/(3/2) + x^(5/2)/(5/2) + C
= (2/3)x^(3/2) + (2/5)x^(5/2) + C
= (2/3)x√x + (2/5)x²√x + C
Q8Find the following integrals: (i) ∫(1 − x)√x dx (ii) ∫√x(3x² + 2x + 3) dx
(i) ∫(1 − x)√x dx = ∫(x^(1/2) − x^(3/2)) dx:
= (2/3)x^(3/2) − (2/5)x^(5/2) + C
= (2/3)x√x − (2/5)x²√x + C
(ii) ∫√x(3x² + 2x + 3) dx = ∫(3x^(5/2) + 2x^(3/2) + 3x^(1/2)) dx:
= 3 × x^(7/2)/(7/2) + 2 × x^(5/2)/(5/2) + 3 × x^(3/2)/(3/2) + C
= (6/7)x^(7/2) + (4/5)x^(5/2) + 2x^(3/2) + C
= (6/7)x³√x + (4/5)x²√x + 2x√x + C
Q10Find the integral: ∫(√x − 1/√x)² dx.
First expand: (√x − 1/√x)² = x − 2(√x)(1/√x) + 1/x = x − 2 + 1/x.
∫(x − 2 + 1/x) dx = x²/2 − 2x + log|x| + C.
= x²/2 − 2x + log|x| + C
Q14Find the integral: ∫(1 − 1/x²)e^(x + 1/x) dx. [Hint: let x + 1/x = t]
Let t = x + 1/x.
Then dt/dx = 1 − 1/x².
So dt = (1 − 1/x²) dx.
The integral becomes:
∫e^t dt = e^t + C = e^(x + 1/x) + C
Q17∫(2x² + e^x) dx is: (A) x³/3 + e^x + C (B) 2x³/3 + e^x + C (C) 2x³/3 + e^x (D) 2x²/3 + e^x + C
∫(2x² + e^x) dx = 2 × x³/3 + e^x + C = 2x³/3 + e^x + C.
Answer: (B) 2x³/3 + e^x + C
(Note: Option (C) is missing the constant C, so it's incorrect. Integration always needs +C for indefinite integrals.)
Q18∫(1 + x)² / x dx is equal to:
Expand: (1 + x)²/x = (1 + 2x + x²)/x = 1/x + 2 + x.
∫(1/x + 2 + x) dx = log|x| + 2x + x²/2 + C.
= log|x| + 2x + x²/2 + C
Q20Find the integral: ∫(2x − 3cos x + e^x) dx.
∫(2x − 3cos x + e^x) dx
= 2 × x²/2 − 3sin x + e^x + C
= x² − 3sin x + e^x + C
Q22The anti-derivative of (√x + 1/√x) equals: (A) (1/3)x^(1/3) + 2x^(1/2) + C (B) (2/3)x^(3/2) + 2x^(1/2) + C (C) (2/3)x^(3/2) − 2x^(1/2) + C (D) (3/2)x^(3/2) + (1/2)x^(1/2) + C
∫(√x + 1/√x) dx = ∫(x^(1/2) + x^(-1/2)) dx
= x^(3/2)/(3/2) + x^(1/2)/(1/2) + C
= (2/3)x^(3/2) + 2x^(1/2) + C
= (2/3)x^(3/2) + 2√x + C
Answer: (B) (2/3)x^(3/2) + 2x^(1/2) + C
7.6Exercise 7.6 - Integration by Parts
10 key questions shown below covering the ILATE rule and standard patterns.
Q1Find the integral: ∫x sin x dx.
Using integration by parts: ∫u dv = uv − ∫v du.
Using ILATE rule: u = x (algebraic), dv = sin x dx.
du = dx, v = −cos x.
∫x sin x dx = x(−cos x) − ∫(−cos x) dx
= −x cos x + ∫cos x dx
= −x cos x + sin x + C
= sin x − x cos x + C
Q2Find the integral: ∫x sin 3x dx.
By ILATE: u = x, dv = sin 3x dx.
du = dx, v = −cos 3x/3.
∫x sin 3x dx = x(−cos 3x/3) − ∫(−cos 3x/3) dx
= −x cos 3x/3 + (1/3)∫cos 3x dx
= −x cos 3x/3 + (1/3)(sin 3x/3) + C
= −(x cos 3x)/3 + (sin 3x)/9 + C
Q3Find the integral: ∫x² e^x dx.
By ILATE: u = x², dv = e^x dx.
du = 2x dx, v = e^x.
∫x²e^x dx = x²e^x − ∫2xe^x dx
= x²e^x − 2∫xe^x dx
Now find ∫xe^x dx by parts again:
u = x, dv = e^x dx → du = dx, v = e^x.
∫xe^x dx = xe^x − ∫e^x dx = xe^x − e^x.
Substitute back:
= x²e^x − 2(xe^x − e^x) + C
= x²e^x − 2xe^x + 2e^x + C
= e^x(x² − 2x + 2) + C
Q5Find the integral: ∫x log x dx.
By ILATE: u = log x (logarithmic, first priority), dv = x dx.
du = (1/x) dx, v = x²/2.
∫x log x dx = (x²/2) log x − ∫(x²/2)(1/x) dx
= (x²/2) log x − (1/2)∫x dx
= (x²/2) log x − (1/2)(x²/2) + C
= (x²/2) log x − x²/4 + C
Or equivalently: (x²/4)(2 log x − 1) + C
Q7Find the integral: ∫x sin⁻¹(x) dx.
By ILATE: u = sin⁻¹(x) (inverse trig, first priority), dv = x dx.
du = 1/√(1−x²) dx, v = x²/2.
∫x sin⁻¹(x) dx = (x²/2)sin⁻¹(x) − ∫(x²/2) × 1/√(1−x²) dx
= (x²/2)sin⁻¹(x) − (1/2)∫x²/√(1−x²) dx
For ∫x²/√(1−x²) dx, substitute x = sin t:
= ∫sin²t/cos t × cos t dt = ∫sin²t dt = ∫(1−cos 2t)/2 dt
= t/2 − sin 2t/4 = t/2 − (sin t cos t)/2
= sin⁻¹(x)/2 − x√(1−x²)/2.
So: ∫x sin⁻¹(x) dx = (x²/2)sin⁻¹(x) − (1/2)[sin⁻¹(x)/2 − x√(1−x²)/2] + C
= (x²/2)sin⁻¹(x) − sin⁻¹(x)/4 + x√(1−x²)/4 + C
= ((2x²−1)/4)sin⁻¹(x) + x√(1−x²)/4 + C
Q10Find the integral: ∫(sin⁻¹x)² dx.
By parts: u = (sin⁻¹x)², dv = dx.
du = 2sin⁻¹(x)/√(1−x²) dx, v = x.
∫(sin⁻¹x)² dx = x(sin⁻¹x)² − ∫2x sin⁻¹(x)/√(1−x²) dx
For the remaining integral, let sin⁻¹(x) = t, so x = sin t, dx = cos t dt.
2x sin⁻¹(x)/√(1−x²) dx = 2(sin t)(t)/cos t × cos t dt = 2t sin t dt.
∫2t sin t dt (by parts): = 2[−t cos t + sin t] = −2t cos t + 2sin t.
Back-substitute: t = sin⁻¹(x), cos t = √(1−x²), sin t = x.
= −2sin⁻¹(x)√(1−x²) + 2x.
Final answer: x(sin⁻¹x)² − [−2sin⁻¹(x)√(1−x²) + 2x] + C
= x(sin⁻¹x)² + 2√(1−x²)sin⁻¹(x) − 2x + C
Q11Find the integral: ∫x cos⁻¹(x) dx. [Hint: compare with Q7]
By ILATE: u = cos⁻¹(x), dv = x dx.
du = −1/√(1−x²) dx, v = x²/2.
∫x cos⁻¹(x) dx = (x²/2)cos⁻¹(x) − ∫(x²/2)(−1/√(1−x²)) dx
= (x²/2)cos⁻¹(x) + (1/2)∫x²/√(1−x²) dx
From Q7, we know ∫x²/√(1−x²) dx = sin⁻¹(x)/2 − x√(1−x²)/2.
Substituting:
= (x²/2)cos⁻¹(x) + (1/2)[sin⁻¹(x)/2 − x√(1−x²)/2] + C
= (x²/2)cos⁻¹(x) + sin⁻¹(x)/4 − x√(1−x²)/4 + C.
Using sin⁻¹(x) = π/2 − cos⁻¹(x) to express in terms of cos⁻¹(x):
= (x²/2)cos⁻¹(x) + (π/2 − cos⁻¹(x))/4 − x√(1−x²)/4 + C
= (x²/2)cos⁻¹(x) − cos⁻¹(x)/4 − x√(1−x²)/4 + C' (absorbing π/8 into C')
= ((2x²−1)/4)cos⁻¹(x) − x√(1−x²)/4 + C.
Q15Find the integral: ∫e^x(sin x + cos x) dx.
This is of the form ∫e^x[f(x) + f'(x)] dx = e^x f(x) + C.
Here f(x) = sin x and f'(x) = cos x. ✓
∫e^x(sin x + cos x) dx = e^x sin x + C
Key formula: ∫e^x[f(x) + f'(x)] dx = e^x f(x) + C.
This is one of the most important integration formulas for CBSE board exams.
Q21Find the integral: ∫e^(2x) sin x dx.
Let I = ∫e^(2x) sin x dx.
By parts: u = sin x, dv = e^(2x) dx.
du = cos x dx, v = e^(2x)/2.
I = (e^(2x) sin x)/2 − ∫(e^(2x)/2) cos x dx
= (e^(2x) sin x)/2 − (1/2)∫e^(2x) cos x dx
Apply by parts again to ∫e^(2x) cos x dx:
u = cos x, dv = e^(2x) dx.
du = −sin x dx, v = e^(2x)/2.
∫e^(2x) cos x dx = (e^(2x) cos x)/2 − ∫(e^(2x)/2)(−sin x) dx
= (e^(2x) cos x)/2 + (1/2)I
Substitute back:
I = (e^(2x) sin x)/2 − (1/2)[(e^(2x) cos x)/2 + (1/2)I]
= (e^(2x) sin x)/2 − (e^(2x) cos x)/4 − I/4
I + I/4 = (e^(2x) sin x)/2 − (e^(2x) cos x)/4
(5/4)I = e^(2x)(2sin x − cos x)/4
I = e^(2x)(2sin x − cos x)/5 + C
Q24∫e^x sec x(1 + tan x) dx equals: (A) e^x cos x + C (B) e^x sec x + C (C) e^x sin x + C (D) e^x tan x + C
∫e^x sec x(1 + tan x) dx = ∫e^x(sec x + sec x tan x) dx.
This is of the form ∫e^x[f(x) + f'(x)] dx where:
f(x) = sec x and f'(x) = sec x tan x. ✓
Using the formula: ∫e^x[f(x) + f'(x)] dx = e^x f(x) + C.
= e^x sec x + C
Answer: (B) e^x sec x + C
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Ask AI TutorPrevious Year Board Exam Questions
Evaluate: ∫e^x(1 + sin x)/(1 + cos x) dx.
Rewrite: (1 + sin x)/(1 + cos x).
1 + cos x = 2cos²(x/2) and 1 + sin x = 1 + 2sin(x/2)cos(x/2).
So (1 + sin x)/(1 + cos x) = [1 + 2sin(x/2)cos(x/2)] / [2cos²(x/2)]
= 1/(2cos²(x/2)) + 2sin(x/2)cos(x/2)/(2cos²(x/2))
= (1/2)sec²(x/2) + tan(x/2).
So the integral becomes:
∫e^x[tan(x/2) + (1/2)sec²(x/2)] dx.
This is of the form ∫e^x[f(x) + f'(x)] dx where f(x) = tan(x/2) and f'(x) = (1/2)sec²(x/2). ✓
= e^x tan(x/2) + C
Evaluate: ∫(0 to π/2) log(sin x) dx.
Let I = ∫(0 to π/2) log(sin x) dx.
Using the property ∫(0 to a) f(x) dx = ∫(0 to a) f(a − x) dx:
I = ∫(0 to π/2) log(sin(π/2 − x)) dx = ∫(0 to π/2) log(cos x) dx.
Add: 2I = ∫(0 to π/2) [log(sin x) + log(cos x)] dx
= ∫(0 to π/2) log(sin x cos x) dx
= ∫(0 to π/2) log(sin 2x / 2) dx
= ∫(0 to π/2) log(sin 2x) dx − ∫(0 to π/2) log 2 dx.
For ∫(0 to π/2) log(sin 2x) dx, let 2x = t, dx = dt/2:
= (1/2)∫(0 to π) log(sin t) dt = (1/2) × 2∫(0 to π/2) log(sin t) dt = I.
(Using the property ∫(0 to 2a) f(x) dx = 2∫(0 to a) f(x) dx when f(2a − x) = f(x), since sin(π − t) = sin t.)
So: 2I = I − (π/2)log 2.
I = −(π/2)log 2.
Find: ∫(2x + 3)/(x² + 3x + 5) dx.
Notice that d/dx(x² + 3x + 5) = 2x + 3.
So the numerator is exactly the derivative of the denominator.
∫(2x + 3)/(x² + 3x + 5) dx = log|x² + 3x + 5| + C.
(This is of the form ∫f'(x)/f(x) dx = log|f(x)| + C.)
Evaluate: ∫x² e^(x³) dx.
Let t = x³, then dt = 3x² dx → x² dx = dt/3.
∫x² e^(x³) dx = ∫e^t (dt/3) = (1/3)e^t + C = (1/3)e^(x³) + C.
Evaluate: ∫e^x(x − 1)/x² dx.
Rewrite: (x − 1)/x² = 1/x − 1/x².
∫e^x(1/x − 1/x²) dx.
Let f(x) = 1/x, then f'(x) = −1/x².
So the integral is of the form ∫e^x[f(x) + f'(x)] dx = e^x f(x) + C.
= e^x/x + C = e^x · (1/x) + C.
Evaluate: ∫(0 to π) x sin x/(1 + cos²x) dx.
Let I = ∫(0 to π) x sin x/(1 + cos²x) dx.
Using the property ∫(0 to a) f(x) dx = ∫(0 to a) f(a − x) dx:
I = ∫(0 to π) (π − x)sin(π − x)/(1 + cos²(π − x)) dx
= ∫(0 to π) (π − x)sin x/(1 + cos²x) dx.
Add: 2I = ∫(0 to π) π sin x/(1 + cos²x) dx.
I = (π/2)∫(0 to π) sin x/(1 + cos²x) dx.
Let t = cos x, dt = −sin x dx.
When x = 0: t = 1; when x = π: t = −1.
I = (π/2)∫(1 to −1) −dt/(1 + t²) = (π/2)∫(−1 to 1) dt/(1 + t²)
= (π/2)[tan⁻¹(t)](from −1 to 1)
= (π/2)[tan⁻¹(1) − tan⁻¹(−1)]
= (π/2)[π/4 − (−π/4)]
= (π/2)(π/2)
= π²/4.
Find: ∫dx/(x² + 4x + 8).
Complete the square: x² + 4x + 8 = (x + 2)² + 4 = (x + 2)² + 2².
∫dx/[(x + 2)² + 2²] = (1/2)tan⁻¹[(x + 2)/2] + C.
= (1/2)tan⁻¹((x + 2)/2) + C.
(Using the formula ∫dx/(x² + a²) = (1/a)tan⁻¹(x/a) + C.)
Evaluate: ∫x sin⁻¹(x)/√(1 − x²) dx.
Let t = sin⁻¹(x), then x = sin t and dx = cos t dt.
Also √(1 − x²) = √(1 − sin²t) = cos t.
∫(sin t)(t)/cos t × cos t dt = ∫t sin t dt.
By parts: u = t, dv = sin t dt → du = dt, v = −cos t.
= −t cos t + ∫cos t dt
= −t cos t + sin t + C.
Back-substitute: t = sin⁻¹(x), cos t = √(1 − x²), sin t = x.
= −sin⁻¹(x)√(1 − x²) + x + C.
Evaluate: ∫(0 to 1) x(1 − x)^n dx.
Using the property ∫(0 to a) f(x) dx = ∫(0 to a) f(a − x) dx:
∫(0 to 1) x(1 − x)^n dx = ∫(0 to 1) (1 − x)(x)^n dx = ∫(0 to 1) (x^n − x^(n+1)) dx.
= [x^(n+1)/(n+1) − x^(n+2)/(n+2)](from 0 to 1)
= 1/(n+1) − 1/(n+2)
= [(n+2) − (n+1)] / [(n+1)(n+2)]
= 1/[(n+1)(n+2)].
Find: ∫(3x + 5)/(x² + 3x − 18) dx.
Factor denominator: x² + 3x − 18 = (x + 6)(x − 3).
Partial fractions: (3x + 5)/[(x + 6)(x − 3)] = A/(x + 6) + B/(x − 3).
3x + 5 = A(x − 3) + B(x + 6).
Put x = 3: 14 = 9B → B = 14/9.
Put x = −6: −13 = −9A → A = 13/9.
∫[(13/9)/(x + 6) + (14/9)/(x − 3)] dx
= (13/9)log|x + 6| + (14/9)log|x − 3| + C
= (1/9)[13 log|x + 6| + 14 log|x − 3|] + C.
Evaluate: ∫(0 to π/2) sin²x/(sin x + cos x) dx.
Let I = ∫(0 to π/2) sin²x/(sin x + cos x) dx.
Using the property: I = ∫(0 to π/2) cos²x/(cos x + sin x) dx.
Add: 2I = ∫(0 to π/2) (sin²x + cos²x)/(sin x + cos x) dx = ∫(0 to π/2) 1/(sin x + cos x) dx.
Now sin x + cos x = √2 sin(x + π/4).
2I = ∫(0 to π/2) dx/[√2 sin(x + π/4)]
= (1/√2)∫(0 to π/2) cosec(x + π/4) dx.
Let u = x + π/4:
= (1/√2)∫(π/4 to 3π/4) cosec u du
= (1/√2)[−log|cosec u + cot u|](from π/4 to 3π/4)
= (1/√2)[−log|cosec(3π/4) + cot(3π/4)| + log|cosec(π/4) + cot(π/4)|]
= (1/√2)[−log|√2 − 1| + log|√2 + 1|]
= (1/√2) log[(√2 + 1)/(√2 − 1)]
= (1/√2) log[(√2 + 1)²] (rationalizing)
= (1/√2) × 2 log(√2 + 1)
= √2 log(√2 + 1)/√2 = (1/√2) log(3 + 2√2).
So I = (1/(2√2)) log(3 + 2√2).
Find: ∫(x² + 1)/(x⁴ + 1) dx.
Divide numerator and denominator by x²:
(1 + 1/x²)/(x² + 1/x²).
Note that x² + 1/x² = (x − 1/x)² + 2.
Let t = x − 1/x, then dt = (1 + 1/x²) dx.
∫dt/(t² + 2) = (1/√2)tan⁻¹(t/√2) + C.
= (1/√2)tan⁻¹[(x² − 1)/(x√2)] + C.
Evaluate: ∫(0 to π/2) (√tan x + √cot x) dx.
∫(0 to π/2) (√tan x + √cot x) dx = ∫(0 to π/2) (√(sin x/cos x) + √(cos x/sin x)) dx
= ∫(0 to π/2) (sin x + cos x)/√(sin x cos x) dx.
Now sin x cos x = (1/2)sin 2x = (1 − (sin x − cos x)²)/2.
Let t = sin x − cos x, dt = (cos x + sin x) dx.
When x = 0: t = −1; when x = π/2: t = 1.
sin x cos x = (1 − t²)/2.
Integral = ∫(−1 to 1) dt/√((1 − t²)/2) = √2 ∫(−1 to 1) dt/√(1 − t²).
= √2 [sin⁻¹(t)](from −1 to 1)
= √2 [π/2 − (−π/2)]
= √2 × π
= π√2.
Evaluate: ∫(1 to 3) (x² + 3x + e^x) dx.
∫(1 to 3) (x² + 3x + e^x) dx = [x³/3 + 3x²/2 + e^x](from 1 to 3)
At x = 3: 27/3 + 27/2 + e³ = 9 + 13.5 + e³ = 22.5 + e³.
At x = 1: 1/3 + 3/2 + e = 1/3 + 3/2 + e = 11/6 + e.
= (22.5 + e³) − (11/6 + e)
= 22.5 − 11/6 + e³ − e
= 135/6 − 11/6 + e³ − e
= 124/6 + e³ − e = 62/3 + e³ − e.
Want more PYQ-style practice? AI generates unlimited similar questions from this chapter.
Practice PYQ PatternsAssertion-Reason Questions
(A) Both true, Reason explains Assertion · (B) Both true, Reason doesn't explain · (C) Assertion true, Reason false · (D) Assertion false, Reason true
AR1
Assertion (A): ∫sin²x dx = x/2 − sin(2x)/4 + C.
Reason (R): sin²x = (1 − cos 2x)/2.
Answer: (A)
Both are TRUE and the reason correctly explains the assertion. Using sin²x = (1 − cos 2x)/2, we get ∫sin²x dx = ∫(1 − cos 2x)/2 dx = x/2 − sin(2x)/4 + C.
AR2
Assertion (A): ∫e^x(sin x + cos x) dx = e^x sin x + C.
Reason (R): ∫e^x[f(x) + f'(x)] dx = e^x f(x) + C.
Answer: (A)
Both are TRUE and the reason correctly explains the assertion. Here f(x) = sin x and f'(x) = cos x. Applying the formula ∫e^x[f(x) + f'(x)] dx = e^x f(x) + C gives e^x sin x + C.
AR3
Assertion (A): ∫(−1 to 1) x³ dx = 0.
Reason (R): x³ is an odd function, i.e., f(−x) = −f(x).
Answer: (A)
Both are TRUE and the reason correctly explains the assertion. Since x³ is an odd function (f(−x) = (−x)³ = −x³ = −f(x)), and the integral is over a symmetric interval [−1, 1], the result is 0 by the property ∫(−a to a) f(x) dx = 0 for odd f.
AR4
Assertion (A): ∫(0 to π) sin³x dx = 4/3.
Reason (R): sin³x = sin x(1 − cos²x) and cos(π − x) = −cos x.
Answer: (A)
Both are TRUE and the reason explains the assertion. Since sin(π − x) = sin x, the function sin³x satisfies f(π − x) = f(x), so ∫(0 to π) sin³x dx = 2∫(0 to π/2) sin³x dx. Using the reduction formula or sin³x = sin x − sin x cos²x: 2∫(0 to π/2) (sin x − sin x cos²x) dx = 2[−cos x + cos³x/3](from 0 to π/2) = 2[(0 + 0) − (−1 + 1/3)] = 2[2/3] = 4/3.
AR5
Assertion (A): ∫1/(x² + 4) dx = (1/2)log|(x² + 4)| + C.
Reason (R): ∫1/(x² + a²) dx = (1/a)tan⁻¹(x/a) + C.
Answer: (D)
Assertion is FALSE: The correct integral is (1/2)tan⁻¹(x/2) + C (using the formula with a = 2), not a logarithmic form. The log form would apply to ∫f'(x)/f(x) dx, but 1/(x² + 4) is not of that form (the derivative of x² + 4 is 2x, not 1). Reason is TRUE: The formula ∫1/(x² + a²) dx = (1/a)tan⁻¹(x/a) + C is correct.
AR6
Assertion (A): ∫(0 to 2a) f(x) dx = 2∫(0 to a) f(x) dx when f(2a − x) = f(x).
Reason (R): ∫(0 to 2a) f(x) dx = ∫(0 to a) f(x) dx + ∫(0 to a) f(2a − x) dx.
Answer: (A)
Both are TRUE and the reason correctly explains the assertion. We split ∫(0 to 2a) into ∫(0 to a) + ∫(a to 2a). In the second integral, let t = 2a − x to get ∫(0 to a) f(2a − t) dt. When f(2a − x) = f(x), this equals ∫(0 to a) f(x) dx. Adding both gives 2∫(0 to a) f(x) dx.
Case Study Questions
A reservoir is being filled by a pipe. The rate of water flow into the reservoir at time t hours is given by R(t) = 6t − t² (in thousand litres per hour), where 0 ≤ t ≤ 6. The reservoir was empty at t = 0.
(i) Find the total amount of water that flows into the reservoir from t = 0 to t = 6 hours. (2 marks)
Total water = ∫(0 to 6) R(t) dt = ∫(0 to 6) (6t − t²) dt
= [6t²/2 − t³/3](from 0 to 6)
= [3t² − t³/3](from 0 to 6)
= 3(36) − 216/3 − 0
= 108 − 72
= 36 thousand litres.
(ii) At what time t is the rate of flow maximum? What is the maximum rate? (2 marks)
R(t) = 6t − t². R'(t) = 6 − 2t.
Set R'(t) = 0: t = 3 hours.
R''(t) = −2 < 0 → maximum confirmed.
Maximum rate = R(3) = 6(3) − 9 = 18 − 9 = 9 thousand litres per hour at t = 3 hours.
(iii)(a) Find the amount of water that flows into the reservoir from t = 0 to t = 3 hours. (2 marks)
∫(0 to 3) (6t − t²) dt = [3t² − t³/3](from 0 to 3)
= 3(9) − 27/3
= 27 − 9
= 18 thousand litres.
Note: This is exactly half of the total flow (36 thousand litres), which makes sense because the flow rate function is symmetric about t = 3 on the interval [0, 6].
(iii)(b) For what value of T (0 < T < 6) does ∫(0 to T) R(t) dt = 18 (half the total flow)? (2 marks)
∫(0 to T) (6t − t²) dt = 18.
[3t² − t³/3](from 0 to T) = 18.
3T² − T³/3 = 18.
Multiply by 3: 9T² − T³ = 54.
T³ − 9T² + 54 = 0.
By inspection, T = 3 is a root: 27 − 81 + 54 = 0. ✓
Factor: (T − 3)(T² − 6T − 18) = 0.
The other roots T = 3 ± 3√3. Since 3 + 3√3 ≈ 8.2 > 6 and 3 − 3√3 < 0, the only valid solution is T = 3 hours.
A manufacturing company finds that the marginal cost of producing x units of a product is given by MC = 3x² − 14x + 15 (in rupees). The fixed cost of production is Rs 50 (cost when x = 0). The selling price per unit is Rs 12.
(i) Find the total cost function C(x). (2 marks)
C(x) = ∫MC dx + Fixed Cost = ∫(3x² − 14x + 15) dx + 50
= x³ − 7x² + 15x + k.
When x = 0: C(0) = k = 50 (fixed cost).
C(x) = x³ − 7x² + 15x + 50.
(ii) Find the cost of producing 5 units. (1 mark)
C(5) = 125 − 7(25) + 15(5) + 50
= 125 − 175 + 75 + 50
= 75 rupees.
(iii)(a) Find the profit function P(x) and determine the number of units for which the profit is maximum. (2 marks)
Revenue R(x) = 12x (selling price × quantity).
Profit P(x) = R(x) − C(x) = 12x − (x³ − 7x² + 15x + 50)
= −x³ + 7x² − 3x − 50.
P'(x) = −3x² + 14x − 3 = 0.
3x² − 14x + 3 = 0.
Discriminant = 196 − 36 = 160. √160 = 4√10 ≈ 12.65.
x = (14 ± 4√10)/6.
x ≈ (14 + 12.65)/6 ≈ 4.4 or x ≈ (14 − 12.65)/6 ≈ 0.2.
Since x must be a positive integer, check x = 4 and x = 5:
P(4) = −64 + 112 − 12 − 50 = −14.
P(5) = −125 + 175 − 15 − 50 = −15.
P''(x) = −6x + 14. P''(4.4) = −26.4 + 14 = −12.4 < 0 → maximum confirmed.
Maximum profit is at x = 4 units (P = −14, still a loss at this selling price).
(iii)(b) Find the average cost of producing x units. For what value of x is the average cost minimum? (2 marks)
Average cost AC(x) = C(x)/x = x² − 7x + 15 + 50/x.
d(AC)/dx = 2x − 7 − 50/x².
Set d(AC)/dx = 0: 2x − 7 − 50/x² = 0.
2x³ − 7x² − 50 = 0.
By inspection, x = 5: 2(125) − 7(25) − 50 = 250 − 175 − 50 = 25 ≠ 0.
Trying x ≈ 4.5: 2(91.125) − 7(20.25) − 50 = 182.25 − 141.75 − 50 = −9.5.
Trying x ≈ 5.2: 2(140.6) − 7(27.04) − 50 = 281.2 − 189.28 − 50 = 41.92.
So the root is between 4.5 and 5.2. For CBSE purposes, the minimum average cost occurs at approximately x = 5 units.
AC(5) = 25 − 35 + 15 + 10 = 15 rupees per unit.
MCQ Practice
MCQ1. ∫1/(1 + x²) dx is equal to:
(A) tan⁻¹x + C
(B) sin⁻¹x + C
(C) cos⁻¹x + C
(D) sec⁻¹x + C
Answer: (A) - ∫1/(1 + x²) dx = **tan⁻¹x + C**. This is a standard integral formula.
MCQ2. ∫e^x(1/x − 1/x²) dx is equal to:
(A) e^x/x² + C
(B) e^x/x + C
(C) e^x·x + C
(D) −e^x/x + C
Answer: (B) - This is of the form ∫e^x[f(x) + f'(x)] dx with f(x) = 1/x and f'(x) = −1/x². So the answer is e^x · f(x) = **e^x/x + C**.
MCQ3. ∫sin²x cos³x dx is equal to:
(A) sin³x/3 − sin⁵x/5 + C
(B) cos³x/3 − cos⁵x/5 + C
(C) sin³x/3 + sin⁵x/5 + C
(D) cos³x/3 + cos⁵x/5 + C
Answer: (A) - ∫sin²x cos³x dx = ∫sin²x cos²x cos x dx = ∫sin²x(1 − sin²x) cos x dx. Let t = sin x, dt = cos x dx. ∫t²(1 − t²) dt = t³/3 − t⁵/5 + C = **sin³x/3 − sin⁵x/5 + C**.
MCQ4. ∫dx/√(9 − x²) is equal to:
(A) sin⁻¹(x/3) + C
(B) cos⁻¹(x/3) + C
(C) tan⁻¹(x/3) + C
(D) (1/3)sin⁻¹x + C
Answer: (A) - ∫dx/√(9 − x²) = ∫dx/√(3² − x²) = sin⁻¹(x/3) + C. Using the formula ∫dx/√(a² − x²) = sin⁻¹(x/a) + C with a = 3.
MCQ5. ∫(0 to 1) dx/(1 + x²) is equal to:
(A) π/2
(B) π/3
(C) π/4
(D) π/6
Answer: (C) - [tan⁻¹x](from 0 to 1) = tan⁻¹(1) − tan⁻¹(0) = π/4 − 0 = **π/4**.
MCQ6. ∫(0 to π) sin³x dx is equal to:
(A) 0
(B) 2/3
(C) 4/3
(D) π/3
Answer: (C) - sin³x = sin x(1 − cos²x). Let t = cos x, dt = −sin x dx. Limits: x=0→t=1, x=π→t=−1. = −∫(1 to −1)(1 − t²) dt = ∫(−1 to 1)(1 − t²) dt = [t − t³/3](from −1 to 1) = 2/3 + 2/3 = **4/3**.
MCQ7. If ∫(0 to a) 3x² dx = 8, then the value of a is:
(A) 1
(B) 2
(C) 3
(D) 4
Answer: (B) - ∫(0 to a) 3x² dx = [x³](from 0 to a) = a³. Setting a³ = 8 gives a = **2**.
MCQ8. ∫(0 to π/2) cos²x dx is equal to:
(A) π/2
(B) π/4
(C) π/3
(D) 0
Answer: (B) - cos²x = (1 + cos 2x)/2. ∫(0 to π/2)(1 + cos 2x)/2 dx = (1/2)[x + sin 2x/2](from 0 to π/2) = (1/2)(π/2) = **π/4**.
MCQ9. ∫2/(1 − cos 2x) dx is equal to:
(A) tan x + C
(B) −cot x + C
(C) cot x + C
(D) −tan x + C
Answer: (B) - 1 − cos 2x = 2sin²x. So 2/(1 − cos 2x) = cosec²x. ∫cosec²x dx = **−cot x + C**.
MCQ10. ∫x/(x² + 1) dx is equal to:
(A) log(x² + 1) + C
(B) (1/2)log(x² + 1) + C
(C) tan⁻¹x + C
(D) (1/2)tan⁻¹x + C
Answer: (B) - Let t = x² + 1, dt = 2x dx. ∫(1/t)(dt/2) = (1/2)log|t| + C = **(1/2)log(x² + 1) + C**.
MCQ11. ∫(−π/2 to π/2) sin⁷x dx is equal to:
(A) 0
(B) 2∫(0 to π/2) sin⁷x dx
(C) 1
(D) π/2
Answer: (A) - sin⁷x is an odd function: sin⁷(−x) = −sin⁷x. On symmetric interval [−π/2, π/2], integral of odd function = **0**.
MCQ12. ∫(1 + tan x)/(1 − tan x) dx is equal to:
(A) log|cos x − sin x| + C
(B) −log|cos x − sin x| + C
(C) log|cos x + sin x| + C
(D) −log|cos x + sin x| + C
Answer: (B) - (1 + tan x)/(1 − tan x) = (cos x + sin x)/(cos x − sin x). d/dx(cos x − sin x) = −(sin x + cos x). So ∫ = **−log|cos x − sin x| + C**.
Try Quiz mode - 10 MCQs per topic with instant feedback and scoring.
Start Quiz ModeImportant Questions - Mark-wise
1 Mark Questions
Find: ∫(sin x + cos x) dx.
∫(sin x + cos x) dx = −cos x + sin x + C = **sin x − cos x + C**.
Evaluate: ∫(0 to 1) dx/(1 + x²).
[tan⁻¹x](from 0 to 1) = tan⁻¹(1) − tan⁻¹(0) = π/4 − 0 = **π/4**.
Find ∫e^x(sin x + cos x) dx.
Using ∫e^x[f(x) + f'(x)] dx = e^x f(x) + C with f(x) = sin x: = **e^x sin x + C**.
∫(2x + e^x) dx is equal to:
∫(2x + e^x) dx = x² + e^x + C.
If d/dx[f(x)] = 4x³ − 3/x⁴ such that f(2) = 0, find f(x).
f(x) = ∫(4x³ − 3x⁻⁴) dx = x⁴ + x⁻³ + C. f(2) = 16 + 1/8 + C = 0 gives C = −129/8. **f(x) = x⁴ + 1/x³ − 129/8**.
2-3 Mark Questions
Find: ∫(2x + 3)/(x² + 3x + 5) dx.
Numerator = derivative of denominator. ∫f'(x)/f(x) dx = log|x² + 3x + 5| + C.
Find: ∫dx/(x² + 4x + 8).
x² + 4x + 8 = (x + 2)² + 2². ∫dx/[(x+2)² + 4] = (1/2)tan⁻¹((x+2)/2) + C.
Find: ∫x log x dx.
By ILATE: u = log x, dv = x dx. = (x²/2)log x − ∫(x/2) dx = (x²/2)log x − x²/4 + C.
Evaluate: ∫(0 to 1) x(1 − x)^n dx.
Using ∫(0 to a) f(x) dx = ∫(0 to a) f(a−x) dx: = ∫(0 to 1)(x^n − x^(n+1)) dx = 1/[(n+1)(n+2)].
3-4 Mark Questions
Find: ∫(3x + 5)/(x² + 3x − 18) dx using partial fractions.
x² + 3x − 18 = (x+6)(x−3). A/(x+6) + B/(x−3). A = 13/9, B = 14/9. = (1/9)[13 log|x+6| + 14 log|x−3|] + C.
Evaluate: ∫e^x(x − 1)/x² dx.
(x−1)/x² = 1/x − 1/x². ∫e^x[f(x) + f'(x)] dx with f(x) = 1/x. = e^x/x + C.
Evaluate: ∫(0 to π) x sin x/(1 + cos²x) dx.
Using f(a−x) property: I = (π/2)∫(0 to π) sin x/(1+cos²x) dx. Substitute t = cos x. = π²/4.
Key Concepts
Important Formulas
Study Plan - Integrals
Basic Integration and Substitution
Read Sections 7.1 and 7.2. Memorize all standard integral formulas (at least 15 formulas). Complete Exercise 7.1 (22 Qs) for basic integrals and at least 20 questions from Exercise 7.2 (39 Qs) on integration by substitution. Focus on recognizing the pattern: when the derivative of the inner function appears as a factor, use substitution.
Trigonometric Integrals, Special Functions, and Partial Fractions
Read Sections 7.3, 7.4, and 7.5. Complete Exercise 7.3 (trig identities, 15+ Qs), Exercise 7.4 (special integral forms, 15+ Qs), and Exercise 7.5 (partial fractions, all 23 Qs). Master the three cases of partial fractions: distinct linear factors, repeated linear factors, and irreducible quadratic factors.
Integration by Parts and Special Types
Read Sections 7.6 and 7.7. Complete Exercise 7.6 (by parts, all 24 Qs) and Exercise 7.7 (11 Qs). Master the ILATE rule, the e^x[f(x)+f'(x)] formula, and the technique for handling repeated integrals (e^x sin x type). These sections carry the most marks in board exams.
Definite Integrals - Fundamental Theorem and Properties
Read Sections 7.8 to 7.11. Complete Exercise 7.8 (limit of sum, 6 Qs), Exercise 7.9 (fundamental theorem, 15+ Qs), Exercise 7.10 (substitution in definite, 10 Qs), and Exercise 7.11 (properties, all 21 Qs). The f(a+b−x) property and odd/even function properties are the most tested. Practice at least 10 problems using these properties.
Miscellaneous Exercise, PYQs, and Full Revision
Complete the Miscellaneous Exercise (at least 25 of 44 Qs). Solve all 14 PYQs from 2021-2026. Revise all formulas (make a formula card). Practice MCQs, assertion-reason, and case study questions. Focus on the top 5 question types: e^x[f+f'] integrals, definite integrals with properties, partial fractions, by parts with inverse trig, and completing the square.
Common Mistakes to Avoid
- ✗Forgetting the constant of integration C in indefinite integrals. Every indefinite integral must include +C. This is a common reason for losing marks in board exams
- ✗Confusing ∫1/(x² + a²) dx = (1/a)tan⁻¹(x/a) + C with ∫1/(x² − a²) dx = (1/2a)log|(x−a)/(x+a)| + C. The sign in the denominator determines whether the answer involves tan⁻¹ or logarithm
- ✗Wrong application of the ILATE rule in integration by parts: choosing the wrong function as u leads to more complicated integrals. Remember: Inverse trig first, then Logarithmic, then Algebraic, then Trig, then Exponential
- ✗In partial fractions, not checking that the degree of the numerator is less than the degree of the denominator. If it is not, perform polynomial long division first before decomposing
- ✗Forgetting to change the limits of integration when making a substitution in definite integrals. If you substitute t = g(x), the limits must change from x = a,b to t = g(a),g(b)
- ✗Misapplying definite integral properties: ∫(0 to 2a) f(x) dx = 2∫(0 to a) f(x) dx ONLY when f(2a−x) = f(x). If f(2a−x) = −f(x), the integral equals 0. Always verify the condition before using the property
Tips for Board Exam
- ✓Master the ∫e^x[f(x) + f'(x)] = e^x f(x) + C formula. It appears in almost every CBSE board paper. Whenever you see e^x multiplied by a sum of two functions, check if one is the derivative of the other
- ✓For definite integrals using properties (5-mark questions), first try the f(a+b−x) substitution. Many integrals simplify dramatically when you replace x with (a+b−x). This is the single most useful technique for 5-mark definite integral problems
- ✓In partial fraction problems, use the cover-up method for speed: to find A in A/(x−a), cover (x−a) in the original fraction and substitute x = a. This is much faster than comparing coefficients
- ✓For integration by parts, if the integral repeats (like ∫e^x sin x dx), let I = the integral and solve the resulting equation for I. This trick appears frequently in board exams
- ✓Time management: allocate about 8 minutes for a 5-mark integration problem. Start with the easier definite integral properties before attempting complex by-parts problems. Always show the substitution and intermediate steps for full marks
Practice Integrals with AI Tutor
Stuck on a question? Ask our AI tutor to explain step-by-step in English or Hindi. Choose from 10 learning modes - Lecture, Quiz, Flashcards, and more. Free for 3 chapters per day.
Start Integrals - Choose Your ModeAbout Integrals - Class 12 Mathematics
Chapter 7 Integrals is the largest and most important chapter in Class 12 Mathematics, covering both indefinite and definite integration. The chapter begins with indefinite integrals (anti-derivatives), establishing that integration is the reverse of differentiation: if d/dx[F(x)] = f(x), then ∫f(x) dx = F(x) + C. Students learn standard integral formulas for polynomial, trigonometric, exponential, and logarithmic functions.
The chapter systematically covers integration techniques: substitution (recognizing composite functions), trigonometric identities (converting powers using half-angle and double-angle formulas), integration of special forms (completing the square for quadratic expressions), partial fractions (decomposing rational functions), and integration by parts using the ILATE rule. The critical formula ∫e^x[f(x) + f'(x)] dx = e^x f(x) + C is highlighted as one of the most tested results.
The second half covers definite integrals: the Fundamental Theorem of Calculus (∫(a to b) f(x) dx = F(b) − F(a)), evaluation as the limit of a sum, and powerful properties including ∫(0 to a) f(x) dx = ∫(0 to a) f(a−x) dx, the odd/even function property on symmetric intervals, and the ∫(0 to 2a) f(x) dx = 2∫(0 to a) f(x) dx property.
For the CBSE 2027 board exam, Integrals typically contributes 9-12 marks (the highest of any single chapter). Students should prioritize: the e^x[f+f'] formula (appears almost every year), definite integrals using properties (5-mark questions), partial fractions (3-mark questions), and integration by parts with inverse trig or log functions. This chapter, combined with Application of Integrals (Chapter 8), forms the core of the 35-mark Calculus unit.
Frequently Asked Questions
How many exercises are in NCERT Class 12 Maths Chapter 7 Integrals?
There are 11 exercises (7.1 to 7.11) plus a Miscellaneous Exercise, totalling approximately 271 questions. This is the largest chapter in the entire Class 12 Maths textbook. Exercise 7.1 has 22 Qs (basic integration), 7.2 has 39 Qs (substitution), 7.3 has 24 Qs (trig identities), 7.4 has 25 Qs (special forms), 7.5 has 23 Qs (partial fractions), 7.6 has 24 Qs (by parts), 7.7 has 11 Qs (special types), 7.8 has 6 Qs (limit of sum), 7.9 has 22 Qs (fundamental theorem), 7.10 has 10 Qs (definite by substitution), 7.11 has 21 Qs (properties), and the Miscellaneous Exercise has 44 Qs.
How many marks does Integrals carry in the CBSE Class 12 board exam?
Integrals is the highest-weightage single chapter. It is part of the Calculus unit (35 marks total). Chapter 7 alone typically contributes 9-12 marks: 1-2 MCQs (1-2 marks), 1 short answer on integration method (3 marks), and 1 long answer on definite integral or integration by parts (5 marks). Combined with Chapter 8 (Application of Integrals), integrals collectively carry about 15-18 marks.
What is the most important formula in Integrals for the board exam?
The formula ∫e^x[f(x) + f'(x)] dx = e^x f(x) + C is arguably the single most important formula. It appears in CBSE board papers almost every year. Master recognizing when a function can be split into f(x) + f'(x) form. Common examples: e^x(sin x + cos x), e^x(1/x − 1/x²), e^x(tan x + sec²x), e^x sec x(1 + tan x).
What is the ILATE rule and how do I use it?
ILATE stands for Inverse trig, Logarithmic, Algebraic, Trigonometric, Exponential. When using integration by parts (∫u dv = uv − ∫v du), choose u as the function that comes first in ILATE order. For example, in ∫x sin x dx, x (Algebraic) comes before sin x (Trigonometric), so u = x. In ∫x log x dx, log x (Logarithmic) comes before x (Algebraic), so u = log x.
What are the key properties of definite integrals for the board exam?
The most important properties are: (1) ∫(a to b) f(x) dx = ∫(a to b) f(a+b−x) dx - used in almost every 5-mark problem. (2) ∫(−a to a) f(x) dx = 0 if f is odd, = 2∫(0 to a) f(x) dx if f is even. (3) ∫(0 to 2a) f(x) dx = 2∫(0 to a) f(x) dx if f(2a−x) = f(x), and = 0 if f(2a−x) = −f(x). (4) ∫(a to b) f(x) dx = ∫(a to c) f(x) dx + ∫(c to b) f(x) dx.
How do I know which integration method to use?
Follow this decision tree: (1) Is it a standard form? Use the formula directly. (2) Can you spot the derivative of an inner function? Use substitution. (3) Is it a rational function with degree of numerator < denominator? Use partial fractions. (4) Is it a product of two different types of functions? Use by parts (ILATE). (5) Does it involve √(a²−x²), √(a²+x²), or √(x²−a²)? Use trig substitution. (6) Is it sin²x, cos²x, or similar? Use double angle identities.
Is NCERT enough for Integrals in the CBSE board exam?
Yes, NCERT is sufficient. All board questions follow NCERT patterns. However, since this is the largest chapter with 271 questions, prioritize: (1) Exercise 7.6 (by parts - most tested), (2) Exercise 7.11 (properties of definite integrals), (3) Exercise 7.5 (partial fractions), (4) Miscellaneous Exercise. Supplement with PYQs (2019-2026) to understand exact patterns. Attempt at least 150 of the 271 questions.
What are the common mistakes students make in Integrals?
The top mistakes are: (1) Forgetting +C in indefinite integrals. (2) Not changing limits when substituting in definite integrals. (3) Confusing tan⁻¹ and log formulas for 1/(x²+a²) vs 1/(x²−a²). (4) Wrong ILATE choice in integration by parts. (5) Not dividing first before partial fractions when numerator degree ≥ denominator degree. (6) Misapplying the f(2a−x) property without verifying the condition.