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JEE • MATHEMATICS • DEFINITE INTEGRATION

Definite integration practice · Easy

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Showing 181-210 of 791 questions

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181 Easy
181 of 791
Understanding +4 | -1
To find the numerical value of ∫−22(px2+qx+s)dx\int\limits_{{-2}}^2\left(p x^2+q x+s\right) d x, it is necessary to know the values of constants
View answer and explanation
Correct option: D
pp and ss
Option D is correct because the integral depends on both p and s, as the integral of px² and s are non-zero over the given interval.
182 Easy
182 of 791
Analyzing +4 | -1
Let I=∫01sin⁡xxdxI=\int\limits_{0}^1 \frac{\sin x}{\sqrt{x}} d x and J=∫01cos⁡xxdxJ=\int\limits_{0}^1 \frac{\cos x}{\sqrt{x}} d x Then which of the following is true?
View answer and explanation
Correct option: B
I<23I<\frac{2}{3} and J<2J<2
The correct answer is B (I<23I<\frac{2}{3} and J<2J<2) because by comparing the integrands with simpler functions and integrating, we can establish the inequalities.
183 Easy
183 of 791
Application +4 | -1
The area bounded by the xx-axis, the curve y=f(x)y=f(x) and the lines x=1,x=bx=1, x=b is equal to b2+1−2\sqrt{b^2+1}-\sqrt{2} for all b>1b>1, then f(x)f(x), is
View answer and explanation
Correct option: D
x/x2+1x / \sqrt{x^2+1}
Option D (x/x2+1x / \sqrt{x^2+1}) is correct because it is obtained by differentiating the given equation with respect to bb using the Fundamental Theorem of Calculus.
184 Easy
184 of 791
Application +4 | -1
The value of the definite integral ∫010((x−5)+(x−5)2+(x−5)3)dx\int\limits_{0}^{10}\left((x-5)+(x-5)^2+(x-5)^3\right) d x is:
View answer and explanation
Correct option: B
2503\frac{250}{3}
Option B (2503\frac{250}{3}) is correct because the integral of the odd powers of (x-5) over a symmetric interval around 5 is zero, and the integral of (x-5)2 is calculated correctly.
185 Easy
185 of 791
Application +4 | -1
The value of ∫01x2exdx\int_0^1 x^2 e^x d x is equal to
View answer and explanation
Correct option: A
e−2e-2
Option A (e−2e-2) is correct because the integral of x2exx^2e^x from 0 to 1 is evaluated using integration by parts twice, resulting in e−2e-2.
186 Easy
186 of 791
Application +4 | -1
Find the value of ∫09[x+2]dx\int\limits_{0}^9[\sqrt{x}+2] d x, where [.] is the greatest integer function
View answer and explanation
Correct option: A
31
Option A (31) is correct because the integral can be evaluated by splitting it into intervals where the greatest integer function has a constant value.
187 Easy
187 of 791
Application +4 | -1
If ∫sin⁡xsin⁡(x−α)dx=Ax+Blog⁡sin⁡(x−α)+C\int \frac{\sin x}{\sin (x-\alpha)} d x=A x+B \log \sin (x-\alpha)+C, then value of (A,B)(A, B) is:
View answer and explanation
Correct option: B
(cos⁡α,sin⁡α)(\cos \alpha, \sin \alpha)
Option B (cos⁡α,sin⁡α\cos \alpha, \sin \alpha) is correct because by substituting x−α=tx - \alpha = t, the integral can be simplified and solved, leading to the coefficients of x and the logarithmic term being cos⁡α\cos \alpha and sin⁡α\sin \alpha respectively.
188 Easy
188 of 791
Application +4 | -1
∫0π/2dx1+(1/6)sin⁡2x=\int\limits_{0}^{\pi / 2} \frac{d x}{1+(1 / 6) \sin ^2 x}=
View answer and explanation
Correct option: A
π267\frac{\pi}{2} \sqrt{\frac{6}{7}}
Option A (π267\frac{\pi}{2} \sqrt{\frac{6}{7}}) is correct because after applying trigonometric identities and integrating, the result is π267\frac{\pi}{2} \sqrt{\frac{6}{7}}.
189 Easy
189 of 791
Application +4 | -1
Let f(x)f(x) be a differentiable function defined on [0,2],\left[ {0,2} \right], such that f′(x)=f′(2−x)f'(x) = f'(2 - x), for all x∈(0,  2), x \in \left( {0,\,\,2} \right),\, f(0)=1f(0) = 1 and f(2)=e2f(2) = {e^2}. Then, the value of ∫02f(x)dx\int\limits_0^2 {f(x)dx} is
View answer and explanation
Correct option: B
1+e21 + {e^2}
Option B (1+e21 + {e^2}) is the correct value of the integral, obtained using the properties of definite integrals and the given conditions.
190 Easy
190 of 791
Application +4 | -1
If ∫0π/2  sin⁡6  x dx=5 π32\int \nolimits_0^{\pi /2} \,\,{\sin ^6}\,\,x\,dx = \frac{{5\,\pi }}{{32}}, then the value of ∫−ππ  (sin⁡6x  +  cos⁡6  x)    dx\int \nolimits_{ - \pi }^\pi \,\,({\sin ^6}x\,\, + \,\,{\cos ^6}\,\,x)\;\,\,dx is
View answer and explanation
Correct option: D
5 π  /  45\,\mathrm{\pi }\,\,/\,\,4
Option D (5 π / 45\,\mathrm{\pi}\,/\,4) is correct because the integral can be simplified using trigonometric identities and the given value of the integral from 0 to π/2\pi/2.
191 Easy
191 of 791
Application +4 | -1
∫−11  cos⁡  h    x1  +  e2x    dx\int \nolimits_{ - 1}^1 \,\,\frac{{\cos \,\,\mathrm{h}\;\,\,x}}{{1\,\, + \,\,{e^{2x}}}}\,\,\;dx is equal to
View answer and explanation
Correct option: C
e2−12e\frac{{{e^2} - 1}}{{2e}}
Option C (e2−12e\frac{{{e^2} - 1}}{{2e}}) is correct because the integral of the given function is solved using properties of hyperbolic cosine and simplification.
192 Easy
192 of 791
Understanding +4 | -1
If (−1,  2)\left( { - 1,\,\,2} \right) and (2,4)\left( {2,4} \right) are two points on the curve y=f(x)y = f\left( x \right) and if g(x)g\left( x \right) is the gradient of the curve at point (x,y)\left( {x,y} \right), then the value of the integral ∫−12  g  (x)  dx\int \limits_{ - 1}^2 \,\,g\,\,\left( x \right)\,\,dx, is
View answer and explanation
Correct option: A
2\mathrm{2}
Option A (2) is correct because the integral of the gradient g(x)g(x) is the difference in the function's values at the limits of integration, f(2)−f(−1)=4−2=2f(2) - f(-1) = 4 - 2 = 2.
193 Easy
193 of 791
Application +4 | -1
If f(x)=sin⁡x∀x∈[0,π2],f(x)+f(π−x)=2∀x∈[π2,π]f(x)=\sin x \forall x \in\left[0, \frac{\pi}{2}\right], f(x)+f(\pi-x)=2 \quad \forall x \in\left[\frac{\pi}{2}, \pi\right] and f(x)=f(2π−x)∀x∈(π,2π]f(x)=f(2 \pi-x) \forall x \in(\pi, 2 \pi]. then the area enclosed by y=f(x)y=f(x) and xx-axis is
View answer and explanation
Correct option: B
2π2 \pi
Option B is correct because the area is calculated by considering the symmetry and the given conditions for f(x).
194 Easy
194 of 791
Application +4 | -1
The value of ∫0π/22sin⁡x2sin⁡x+2cos⁡xdx\int\limits_{0}^{\pi / 2} \frac{2^{\sin x}}{2^{\sin x}+2^{\cos x}} d x is
View answer and explanation
Correct option: A
π4\frac{\pi}{4}
Option A (π4\frac{\pi}{4}) is correct because using the property ∫0af(x)dx=∫0af(a−x)dx\int_0^a f(x) dx = \int_0^a f(a-x) dx simplifies the integral to π4\frac{\pi}{4}.
195 Easy
195 of 791
Analyzing +4 | -1
If ∫01ett+1dt=a\int_0^1 \frac{e^t}{t+1} d t=a, then ∫b−1be−tt−b−1 dt\int_{b-1}^b \frac{e^{-t}}{t-b-1} \,d t is equal to
View answer and explanation
Correct option: B
−ae−b-a e^{-b}
Option B (−ae−b-a e^{-b}) is correct because after substitution and simplification, the integral is equal to −ae−b-a e^{-b}.

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SEQUENCE OF SERIES 1764 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs DIFFERENTIAL EQUATIONS 977 MCQs
196 Easy
196 of 791
Application +4 | -1
If f(x)={ecos⁡xsin⁡x,∣x∣≤22, otherwise f(x)=\left\{\begin{array}{cc}e^{\cos x} \sin x, & |x| \leq 2 \\ 2, & \text { otherwise }\end{array}\right., then ∫−23f(x)dx\int\limits_{{-2}}^3 f(x) d x is equal to
View answer and explanation
Correct option: C
2
Option C (2) is correct because the integral can be split into two parts, one of which is zero due to the odd function property, and the other is a simple integral.
197 Easy
197 of 791
Application +4 | -1
The value of ∫−ππcos⁡2x1+axdx,a>0\int\limits_{{-\pi}}^\pi \frac{\cos ^2 x}{1+a^x} d x, a>0 is
View answer and explanation
Correct option: C
π2\frac{\pi}{2}
Option C is correct because by using the property ∫abf(x)dx=∫abf(a+b−x)dx\int_a^b f(x) dx = \int_a^b f(a+b-x) dx and adding the two integrals, we can simplify the integral to ∫−ππcos⁡2xdx\int_{-\pi}^{\pi} \cos^2 x dx which equals π\pi, and hence I = π2\frac{\pi}{2}.
198 Easy
198 of 791
Application +4 | -1
If 2∫01tan⁡−1x dx=∫01cot⁡−1(1−x+x2)dx.2 \int\limits_{0}^1 \tan ^{-1} x \,d x=\int\limits_{0}^1 \cot ^{-1}\left(1-x+x^2\right) d x. Then, ∫01tan⁡−1(1−x+x2)dx\int\limits_{0}^1 \tan ^{-1}\left(1-x+x^2\right) d x is equal to
View answer and explanation
Correct option: A
log⁡2\log 2
Option A is correct. The equation is simplified using trigonometric identities and integration by parts, leading to the solution log 2.
199 Easy
199 of 791
Application +4 | -1
The value of the integral ∫0100π1−cos⁡2x\int\limits_{0}^{100 \pi} \sqrt{1-\cos 2 x} dx is
View answer and explanation
Correct option: B
2002200 \sqrt{2}
Option B is correct because the integral is solved by using the identity for 1−cos2x\sqrt{1 - cos 2x} and then integrating the absolute value of the sine function over the given interval.
200 Easy
200 of 791
Understanding +4 | -1
To find the numerical value of ∫−22  (px3+qx+s)  dx,  it\int \nolimits_{ - 2}^2 \,\,\left( {p{x^3} + qx + s} \right)\,\,dx,\,\,\mathrm{it} is necessary to know the values of the constants
View answer and explanation
Correct option: C
ss
Option C (s) is correct because the integral of px³ and qx from -2 to 2 is 0 (odd functions), so only the integral of s remains, which depends on the value of s.
201 Easy
201 of 791
Application +4 | -1
The area of the triangle formed by the positive xx-axis and the normal and the tangent to the circle x2+y2x^2+y^2 =4=4 at (1,3)(1, \sqrt{3}) is
View answer and explanation
Correct option: A
232 \sqrt{3}
Option A (232 \sqrt{3}) is correct because the area of the triangle formed by the x-axis, tangent, and normal is 232\sqrt{3}.
202 Easy
202 of 791
Application +4 | -1
The area bounded by the curves y2−x=0y^2-x=0 and y−x2=0y-x^2=0 is
View answer and explanation
Correct option: B
13\frac{1}{3}
Option B (13\frac{1}{3}) is correct because the area bounded by the curves y2−x=0y^2-x=0 and y−x2=0y-x^2=0 is found by integrating the difference of the functions from 0 to 1, which results in 13\frac{1}{3}.
203 Easy
203 of 791
Application +4 | -1
The value of ∫012a  f(x)f(x)  +  f (12 a−x)dx  is\int \nolimits_0^{12a} \,\,\frac{{f\left( x \right)}}{{f\left( x \right)\,\, + \,\,f\,\left( {12\,a - x} \right)}}dx\,\,\mathrm{is}
View answer and explanation
Correct option: D
6 a6\,a
Option D (6a6a) is correct because by using the property of definite integrals, the integral simplifies to half the interval length, which is 6a6a.
204 Easy
204 of 791
Application +4 | -1
∫01(1−x2)3/2dx=\int\limits_{0}^1\left(1-x^2\right)^{3 / 2} d x=
View answer and explanation
Correct option: C
3π16\frac{3 \pi}{16}
Option C (3π16\frac{3 \pi}{16}) is correct because the integral evaluates to 3π16\frac{3 \pi}{16} after trigonometric substitution and simplification.
205 Easy
205 of 791
Analyzing +4 | -1
lim⁡n→∞[1n+n(n+1)2+n(n+2)2+...+n(2n+1)2]\lim\limits_{n \to \infty } \left[ {\frac{1}{n} + \frac{n}{{{{(n + 1)}^2}}} + \frac{n}{{{{(n + 2)}^2}}} + ... + \frac{n}{{{{(2n + 1)}^2}}}} \right] equal to
View answer and explanation
Correct option: B
12 \frac{1}{2}\,
Option B (12 \frac{1}{2}\,) is correct because the limit of the summation can be converted into a definite integral, which evaluates to 12\frac{1}{2}.
206 Easy
206 of 791
Application +4 | -1
The value of ∫−π/2π/21∣x∣+∣sin⁡x]+4dx\int_{-\pi / 2}^{\pi / 2} \frac{1}{|x|+\mid \sin x]+4} d x, where [t][t] denotes the greatest integer less than or equal to t,t, is
View answer and explanation
Correct option: D
320(4π−3)\frac{3}{20}(4 \pi-3)
The correct answer is D (320(4π−3)\frac{3}{20}(4 \pi-3)) because after evaluating the integral using the properties of the greatest integer function, the result is 320(4π−3)\frac{3}{20}(4 \pi-3).
207 Easy
207 of 791
Application +4 | -1
If I=∫−22  ∣1−x4∣  dxI = \int \limits_{ - 2}^2 \,\,\left| {1 - {x^4}} \right|\,\,dx, then   I\mathrm{\;I} equals
View answer and explanation
Correct option: C
12\mathrm{12}
Option C (12) is correct because the integral of the absolute value function |1 - x4| from -2 to 2 is calculated by splitting the integral at the points where 1 - x4 changes sign.
208 Easy
208 of 791
Analyzing +4 | -1
The value of integral ∫−22∣x3+x∣(ex∣x∣+1)dx\int\limits_{ - 2}^2 {\frac{{\left| {{x^3} + x} \right|}}{{({e^{x\left| x \right|}} + 1)}}} dx is equal to:
View answer and explanation
Correct option: D
6
Option D (6) is correct because the integral simplifies to the integral of an even function, which is then easily evaluated.
209 Easy
209 of 791
Analyzing +4 | -1
Let I1=∫01ex1+xdxI_1=\int_0^1 \frac{e^x}{1+x} d x and I2=∫01x2ex3(2−x3)dxI_2=\int_0^1 \frac{x^2}{e^{x^3}\left(2-x^3\right)} d x.
Then, I1I2\frac{I_1}{I_2} is equal to
View answer and explanation
Correct option: C
3e3 e
Option C (3e3e) is correct because after substitution and using properties of definite integrals, the ratio simplifies to 3e3e.
210 Easy
210 of 791
Application +4 | -1
If ∫1x1−x1+xdx=g(x)+c,  g(1)=0,\int {\frac{1}{x}\sqrt {\frac{{1 - x}}{{1 + x}}} dx = g\left( x \right) + c,\,\,g\left( 1 \right) = 0,} then g(12)g\left( {\frac{1}{2}} \right) us equal to
View answer and explanation
Correct option: A
log⁡e(3−13+1)+π3{\log _e}\left( {\frac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}}} \right) + \frac{\pi }{3}
Option A (log⁡e(3−13+1)+π3{\log _e}\left( {\frac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}}} \right) + \frac{\pi }{3}) is correct because it is the result of the definite integral calculation with the given substitution and initial condition.
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