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JEE • MATHEMATICS • DEFINITE INTEGRATION

Definite integration practice · Medium

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301 Medium
301 of 596
Analyzing +4 | -1
Let ff be a continuous function that p(x)=∫−2af(t)∣x−t∣dtp(x)=\int\limits_{{-2}}^a f(t)|x-t| d t where x∈(−2,2)x \in(-2,2), then the value of ∫02p′′(x)f(x)dx\int\limits_{0}^2 \frac{p^{\prime \prime}(x)}{f(x)} d x is equal to
View answer and explanation
Correct option: D
4
Option D (4) is correct because the integral simplifies to 4 after finding the second derivative of p(x) and substituting.
302 Medium
302 of 596
Analyzing +4 | -1
The points of extremum of the function ϕ(x)=∫1xe−t2/2(1−t2)dt\phi \left( x \right) = \int \limits_1^x {e^{ - {t^2}/2}}\left( {1 - {t^2}} \right)dt, are
View answer and explanation
Correct option: B
x=1,  −1x = 1,\; - 1
Option B (x=1,−1x = 1, -1) is correct because these are the points where the derivative of the function equals zero, indicating potential extrema.
303 Medium
303 of 596
Application +4 | -1
The value of ∫π/43π/4ϕ1+sin⁡ϕdϕ\int\limits_{{\pi / 4}}^{3 \pi / 4} \frac{\phi}{1+\sin \phi} d \phi, is
View answer and explanation
Correct option: A
πtan⁡π8\pi \tan \frac{\pi}{8}
Option A (πtan⁡π8\pi \tan \frac{\pi}{8}) is correct because, after applying the property of definite integrals and simplifying, the value of the integral is found to be πtan⁡π8\pi \tan \frac{\pi}{8}.
304 Medium
304 of 596
Analyzing +4 | -1
If I1=∫−4−5e(x+5)2dxI_1=\int\limits_{{-4}}^{-5} e^{(x+5)^2} d x and I2=3∫1/32/3e9(x−(2/3))2dxI_2=3 \int\limits_{{1 / 3}}^{2 / 3} e^{9(x-(2 / 3))^2} d x then the value of I1+I2I_1+I_2 is
View answer and explanation
Correct option: A
0
Option A (0) is correct because after substitution, the two integrals cancel each other out.
305 Medium
305 of 596
Analyzing +4 | -1
If x=∫0ydt1+t2x=\int\limits_{0}^y \frac{d t}{\sqrt{1+t^2}}, then d2ydx2\frac{d^2 y}{d x^2} is equal to:
View answer and explanation
Correct option: A
yy
Option A (y) is correct because it is the result of differentiating the given equation twice and simplifying.
306 Medium
306 of 596
Application +4 | -1
Area bounded by the curves y=sin⁡πx2y=\sin \frac{\pi x}{2} and y=x3y=x^3, is equal to
View answer and explanation
Correct option: D
8−π2π\frac{8-\pi}{2 \pi} sq. units
Option D is correct because it represents the area calculated by integrating the difference between the two curves from 0 to 1, multiplied by 2 due to symmetry.
307 Medium
307 of 596
Application +4 | -1
The value of ∑n=1100∫n−1nex−[x]dx,\sum\limits_{n = 1}^{100} {\int_{n - 1}^n {{e^{x - \left[ x \right]}}dx,} } where [x]\left[ x \right] is the greatest integer ≤x,\le x, is
View answer and explanation
Correct option: A
100 (e−1)100\,(e - 1)
The correct answer is A (100 (e−1)100\,(e - 1)) because it correctly evaluates the integral and uses the properties of the greatest integer function.
308 Medium
308 of 596
Application +4 | -1
lim⁡n→∞∑r=14n1n+r\lim \limits_{{n \rightarrow \infty}} \sum\limits_{{r=1}}^{4 n} \frac{1}{n+r} is
View answer and explanation
Correct option: A
log⁡e5\log _e 5
Option A (log⁡e5\log _e 5) is correct because the limit of the sum can be expressed as a definite integral, which evaluates to log⁡e5\log_e 5.
309 Medium
309 of 596
Application +4 | -1
A function f(x)f(x) satisfies f(x)=f(cx)f(x)=f\left(\frac{c}{x}\right) for some real number c(c>1)c(c>1) and ∀ x>0\forall \, x>0. If ∫1cf(x)xdx=3\int\limits_{1}^{\sqrt{c}} \frac{f(x)}{x} d x=3, then the value of ∫1cf(x)xdx\int\limits_{1}^c \frac{f(x)}{x} d x is
View answer and explanation
Correct option: A
6
Option A (6) is correct because by using the substitution u=cxu = \frac{c}{x} and the given property of the function, the integral from c\sqrt{c} to cc is equal to the integral from 1 to c\sqrt{c}.
310 Medium
310 of 596
Application +4 | -1
∫0π/2xsin⁡xcos⁡xcos⁡4x+sin⁡4xdx=\int\limits_{0}^{\pi / 2} \frac{x \sin x \cos x}{\cos ^4 x+\sin ^4 x} d x=
View answer and explanation
Correct option: D
π216\frac{\pi^2}{16}
Option D (π216\frac{{\pi^2}}{{16}}) is correct because it is the result of evaluating the definite integral using trigonometric substitution and properties of definite integrals.
311 Medium
311 of 596
Application +4 | -1
The value of the integral ∫e−1e2∣ln⁡xx∣dx\int\limits_{e^{-1}}^{e^2}\left|\frac{\ln x}{x}\right| d x is:
View answer and explanation
Correct option: B
52\frac{5}{2}
Option B (52\frac{5}{2}) is correct because the integral is split into two parts based on the sign of ln⁡x\ln x, and then each part is evaluated separately.
312 Medium
312 of 596
Application +4 | -1
For m>0,n>0m>0, n>0, let Im,n=∫01xm(log⁡x)nI_{m, n}=\int\limits_{0}^1 x^m(\log x)^n dx\mathrm{d} x, then I5,5I_{5,5} is given by
View answer and explanation
Correct option: C
−5!66-\frac{5 !}{6^6}
Option C (−5!66-\frac{{5 !}}{{6^6}}) is correct because the integral can be solved by repeated integration by parts.
313 Medium
313 of 596
Analyzing +4 | -1
∫0∞  x  log⁡  x    dx(1+x2)2\int \nolimits_0^\infty \,\,\frac{{x\,\,\log \,\,x\;\,\,dx}}{{{{\left( {1 + {x^2}} \right)}^2}}} is equal to
View answer and explanation
Correct option: A
0\mathrm{0}
Option A (0) is correct because the integral evaluates to zero due to the symmetry of the integrand after a suitable substitution.
314 Medium
314 of 596
Application +4 | -1
∫0ax(2ax−x2)32dx=\int\limits_{0}^a x\left(2 a x-x^2\right)^{\frac{3}{2}} d x=
View answer and explanation
Correct option: C
a5[3π16−15]a^5\left[\frac{3 \pi}{16}-\frac{1}{5}\right]
Option C (a5[3π/16-1/5]) is correct because the integral evaluates to a5[3π/16-1/5] after using trigonometric substitution and simplifying.
315 Medium
315 of 596
Application +4 | -1
Let α>0\alpha > 0. If ∫0αxx+α−xdx=16+20215\int\limits_{{0}}^{\alpha} \frac{x}{\sqrt{x+\alpha}-\sqrt{x}} d x=\frac{16+20 \sqrt{2}}{15}, then α\alpha is equal to:
View answer and explanation
Correct option: C
2\sqrt{2}
Option C (2\sqrt{2}) is correct because it involves solving a definite integral after rationalizing the denominator and then comparing the result with the given value to find α\alpha.

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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
316 Medium
316 of 596
Application +4 | -1
∫04{x}\int\limits_{0}^4\{\sqrt{x}\} is equal to, where {x}\{x\} denotes the fraction part of xx.
View answer and explanation
Correct option: D
73\frac{7}{3}
Option D (73\frac{7}{3}) is correct because the integral of the fractional part of the square root of x from 0 to 4 is equal to 73\frac{7}{3}.
317 Medium
317 of 596
Application +4 | -1
∫01/2  ∣ sin⁡  π  x∣  dx\int \limits_0^{1/2} \,\,\left| {\,\sin \,\,\pi \,\,x} \right|\,\,dx is equal to
View answer and explanation
Correct option: D
1  /  π1\,\,/\,\,\pi
Option D (1/π1/\pi) is correct because the integral is evaluated using substitution and the properties of the sine function.
318 Medium
318 of 596
Application +4 | -1
∫0πxlog⁡sin⁡xdx=\int\limits_{0}^\pi x \log \sin x d x=
View answer and explanation
Correct option: B
π22log⁡12\frac{\pi^2}{2} \log \frac{1}{2}
Option B (π22log⁡12\frac{\pi^2}{2} \log \frac{1}{2}) is correct because the integral can be solved using the properties of definite integrals and trigonometric functions, specifically exploiting the symmetry of the sine function.
319 Medium
319 of 596
Application +4 | -1
∫0π  cos⁡  2x  +  12    dx\int \nolimits_0^\pi \,\,\sqrt {\frac{{\cos \,\,2x\,\, + \,\,1}}{2}} \;\,\,dx is equal to
View answer and explanation
Correct option: B
2\mathrm{2}
Option B (2) is correct. The integral simplifies to ∫0π∣cos⁡x∣dx\int \nolimits_0^{\pi } |\cos x| dx, which evaluates to 2.
320 Medium
320 of 596
Application +4 | -1
The integral ∫sin⁡2xcos⁡2x(sin⁡3x+cos⁡3x)2dx\int \frac{\sin ^2 x \cos ^2 x}{\left(\sin ^3 x+\cos ^3 x\right)^2} d x is equal to:
View answer and explanation
Correct option: B
−13(1+tan⁡3x)+c-\frac{1}{3\left(1+\tan ^3 x\right)}+c
Option B (−13(1+tan⁡3x)+c-\frac{1}{3(1+\tan ^3 x)}+c) is correct because the integral simplifies to this expression after substitution and integration.
321 Medium
321 of 596
Application +4 | -1
∫0π/8  cos⁡3  4  θ  dθ\int \nolimits_0^{\pi /8} \,\,{\cos ^3}\,\,4\,\,\mathrm{\theta }\,\,d\mathrm{\theta } is equal to
View answer and explanation
Correct option: D
1 / 6\mathrm{1}\,\mathrm{/}\,\mathrm{6}
Option D (1 / 6\mathrm{1}\,\mathrm{/}\,\mathrm{6}) is correct because it is the result of evaluating the definite integral of cos⁡3 4 θ{\cos ^3}\,4\,\theta from 0 to π/8\pi/8.
322 Medium
322 of 596
Analyzing +4 | -1
If f(x)=∫(x2+sin⁡2x1+x2)sec⁡2xdxf(x)=\int\left(\frac{x^2+\sin ^2 x}{1+x^2}\right) \sec ^2 x d x and f(0)=0f(0)=0, then f(1)f(1) equals
View answer and explanation
Correct option: A
tan⁡1−π4\tan 1-\frac{\pi}{4}
Option A (tan⁡1−π4\tan 1-\frac{\pi}{4}) is correct because after integrating and using the initial condition f(0)=0, we find f(1) = tan 1 - pi/4.
323 Medium
323 of 596
Application +4 | -1
The value ∫1a[x]f′(x)dx,a>1\int\limits_{1}^a[x] f^{\prime}(x) d x, a > 1, where [x][x] denotes the greatest integer not exceeding xx, is
View answer and explanation
Correct option: B
[a]f(a)−{f(1)+f(2)+…+f([a])}[a] f(a)-\{f(1)+f(2)+\ldots+f([a])\}
The correct answer is B ([a]f(a)−{f(1)+f(2)+…+f([a])}[a] f(a)-\{f(1)+f(2)+\ldots+f([a])\}) because the integral is evaluated by splitting the range and using integration by parts.
324 Medium
324 of 596
Application +4 | -1
∫0π/2log⁡sin⁡xdx=\int\limits_{0}^{\pi / 2} \log \sin x d x=
View answer and explanation
Correct option: A
−(π2)log⁡2-\left(\frac{\pi}{2}\right) \log 2
Option A (−(π2)log⁡2-\left(\frac{\pi}{2}\right) \log 2) is the correct answer because it is the result of evaluating the definite integral using properties of logarithms and trigonometric functions.
325 Medium
325 of 596
Analyzing +4 | -1
Let f(x)f\left( x \right) be an integrable function defined on [a,  b],  b   >   a   >  0\left[ {a,\,\,b} \right],\,\,b\,\,\, > \,\,\,a\,\,\, > \,\,0 If I1=∫π/6π/3  f  (tan⁡  θ  +  cot⁡  θ)sec⁡2  θ  dθ{I_1} = \int \limits_{\pi /6}^{\pi /3} \,\,f\,\,\left( {\tan \,\,\theta \,\, + \,\,\cot \,\,\theta } \right){\sec ^2}\,\,\theta \,\,d\theta and, I2=∫π/6π/3  f(tan⁡  θ  +  cot⁡  θ)  cosec2  θ  dθ{I_2} = \int \limits_{\pi /6}^{\pi /3} \,\,f\left( {\tan \,\,\theta \,\, + \,\,\cot \,\,\theta } \right)\,\,\mathrm{cose}{\mathrm{c}^2}\,\,\theta \,\,d\theta, then I1I2=\frac{{{I_1}}}{{{I_2}}} =
View answer and explanation
Correct option: A
A positive integer
Option A (A positive integer) is correct because the integral I1 equals I2, making the ratio equal to 1, which is a positive integer.
326 Medium
326 of 596
Application +4 | -1
∫03  ∣x3  +  x2  +  3x∣dx\int \nolimits_0^3 \,\,\left| {{x^3}\,\, + \,\,{x^2}\,\, + \,\,3x} \right|dx is equal to
View answer and explanation
Correct option: B
1714\frac{{171}}{4}
Option B (1714\frac{{171}}{4}) is correct because the integrand is always non-negative in the interval [0, 3], so the absolute value can be dropped, and the resulting polynomial can be integrated directly.
327 Medium
327 of 596
Application +4 | -1
The values of the integral ∫e−1e2∣log⁡exx∣dx\int\limits_{e^{-1}}^{e^2}\left|\frac{\log _e x}{x}\right| d x is
View answer and explanation
Correct option: A
32\frac{3}{2}
Option A (32\frac{{3}}{{2}}) is correct. The integral is solved by splitting the integral at x=1 due to the absolute value and then evaluating each part.
328 Medium
328 of 596
Application +4 | -1
∫−11  [x  sin⁡  πx]  dx\int \nolimits_{ - 1}^1 \,\,[x\,\,\sin \,\,\mathrm{\pi }x]\,\,dx is equal to
View answer and explanation
Correct option: D
0\mathrm{0}
Option D (0) is correct because the integrand is an even function and the value of the function inside the integral is zero within the integration limits.
329 Medium
329 of 596
Analyzing +4 | -1
Let f(x)=x2+ax+bf(x)=x^2+a x+b and the only solution of the equation f(x)=f(x)= minimum f(x)f(x) is x=0x=0 and f(x)=0f(x)=0 has root α\alpha and β\beta, then ∫αβx3dx\int_\alpha^\beta x^3 d x is equal to
View answer and explanation
Correct option: C
0
Option C (0) is correct because the integral of an odd function over a symmetric interval is zero.
330 Medium
330 of 596
Analyzing +4 | -1
Let In=∫0π/2  sin⁡n  x  dx,  n∈N{I_n} = \int \limits_0^{\pi /2} \,\,{\sin ^n}\,\,x\,\,dx,\,\,n \in N. Then
View answer and explanation
Correct option: C
n (In−2−In)=In−2n\,\left( {{I_{n - 2}} - {I_n}} \right) = {I_{n - 2}}
Option C (n (In−2−In)=In−2n\,\left( {{I_{n - 2}} - {I_n}} \right) = {I_{n - 2}}) is correct because it represents the correct reduction formula derived from integration by parts.
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