Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Definite Integration question

2021 · 22 Jul · Shift 2 · Q28
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Definite Integration
  5. /2021 · 22 Jul · Shift 2 · Q28

Definite Integration question

2021 · 22 Jul · Shift 2 · Q28

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1
If ∫0100πsin⁡2xe(xπ−[xπ])dx=απ31+4π2,α∈R\int\limits_0^{100\pi } {{{{{\sin }^2}x} \over {{e^{\left( {{x \over \pi } - \left[ {{x \over \pi }} \right]} \right)}}}}dx = {{\alpha {\pi ^3}} \over {1 + 4{\pi ^2}}},\alpha \in R}0∫100π​e(πx​−[πx​])sin2x​dx=1+4π2απ3​,α∈R where [x] is the greatest integer less than or equal to x, then the value of α\alphaα is :
  1. A
    200 (1 −-− e −-− 1)
  2. B
    100 (1 −-− e)
  3. C
    50 (e −-− 1)
  4. D
    150 (e −-− 1 −-− 1)
View written solutionFree

Correct answer: A

  1. Rewrite the integrand using the fractional part

    Given I=∫0100πsin⁡2xe(xπ−[xπ]) dx.I=\int_0^{100\pi}\frac{\sin^2 x}{e^{\left(\frac{x}{\pi}-\left[\frac{x}{\pi}\right]\right)}}\,dx.I=∫0100π​e(πx​−[πx​])sin2x​dx.

    Note that xπ−[xπ]={xπ},\frac{x}{\pi}-\left[\frac{x}{\pi}\right] = \left\{\frac{x}{\pi}\right\},πx​−[πx​]={πx​}, the fractional part of xπ\frac{x}{\pi}πx​.

    So on each interval x∈[nπ,(n+1)π),n=0,1,2,…,99,x\in[n\pi,(n+1)\pi),\quad n=0,1,2,\dots,99,x∈[nπ,(n+1)π),n=0,1,2,…,99, we have [xπ]=n,\left[\frac{x}{\pi}\right]=n,[πx​]=n, hence xπ−[xπ]=xπ−n.\frac{x}{\pi}-\left[\frac{x}{\pi}\right]=\frac{x}{\pi}-n.πx​−[πx​]=πx​−n.

  2. Break the integral into 100 equal parts

    Therefore, I=∑n=099∫nπ(n+1)πsin⁡2x e−(xπ−n)dx.I=\sum_{n=0}^{99}\int_{n\pi}^{(n+1)\pi}\sin^2 x\,e^{-\left(\frac{x}{\pi}-n\right)}dx.I=∑n=099​∫nπ(n+1)π​sin2xe−(πx​−n)dx.

    Put x=nπ+t,0≤t≤π,dx=dt.x=n\pi+t,\quad 0\le t\le \pi,\quad dx=dt.x=nπ+t,0≤t≤π,dx=dt.

    Then sin⁡2(nπ+t)=sin⁡2t\sin^2(n\pi+t)=\sin^2 tsin2(nπ+t)=sin2t and e−(nπ+tπ−n)=e−t/π.e^{-\left(\frac{n\pi+t}{\pi}-n\right)}=e^{-t/\pi}.e−(πnπ+t​−n)=e−t/π.

    So each sub-integral is the same:

    =\int_0^{\pi}\sin^2 t\,e^{-t/\pi}dt.$$ Hence $$I=100\int_0^{\pi}\sin^2 t\,e^{-t/\pi}dt.$$

  3. Use the identity for sin⁡2t\sin^2 tsin2t

    sin⁡2t=1−cos⁡2t2.\sin^2 t=\frac{1-\cos 2t}{2}.sin2t=21−cos2t​.

    Thus I=100⋅12(∫0πe−t/πdt−∫0πe−t/πcos⁡2t dt).I=100\cdot \frac12\left(\int_0^{\pi}e^{-t/\pi}dt-\int_0^{\pi}e^{-t/\pi}\cos 2t\,dt\right).I=100⋅21​(∫0π​e−t/πdt−∫0π​e−t/πcos2tdt).

  4. Evaluate the first integral

    =\left[-\pi e^{-t/\pi}\right]_0^{\pi} =\pi(1-e^{-1}).$$

  5. Evaluate the second integral

    Use the standard result ∫eatcos⁡bt dt=eata2+b2(acos⁡bt+bsin⁡bt).\int e^{at}\cos bt\,dt=\frac{e^{at}}{a^2+b^2}(a\cos bt+b\sin bt).∫eatcosbtdt=a2+b2eat​(acosbt+bsinbt).

    Here a=−1π,b=2.a=-\frac1\pi,\qquad b=2.a=−π1​,b=2.

    So

    =\left[\frac{e^{-t/\pi}}{\frac1{\pi^2}+4}\left(-\frac1\pi\cos 2t+2\sin 2t\right)\right]_0^{\pi}.$$ Since $$\cos 2\pi=1,\ \sin 2\pi=0,\ \cos 0=1,\ \sin 0=0,$$ we get $$\int_0^{\pi}e^{-t/\pi}\cos 2t\,dt =\frac{-\frac{e^{-1}}\pi+\frac1\pi}{\frac1{\pi^2}+4}.

    Simplify:

    =\frac{\frac{1-e^{-1}}\pi}{\frac{1+4\pi^2}{\pi^2}} =\frac{\pi(1-e^{-1})}{1+4\pi^2}.$$

  6. Substitute back

    Therefore, I=50(π(1−e−1)−π(1−e−1)1+4π2).I=50\left(\pi(1-e^{-1})-\frac{\pi(1-e^{-1})}{1+4\pi^2}\right).I=50(π(1−e−1)−1+4π2π(1−e−1)​).

    Factor:

    =50\pi(1-e^{-1})\cdot\frac{4\pi^2}{1+4\pi^2}.$$ Hence $$I=\frac{200\pi^3(1-e^{-1})}{1+4\pi^2}.$$ Comparing with $$I=\frac{\alpha\pi^3}{1+4\pi^2},$$ we get $$\alpha=200(1-e^{-1}).$$

  7. Match with the options

    Option A is intended as 200(1−e−1).200(1-e^{-1}).200(1−e−1).

    So the correct choice is A.

  8. Compare with stored correct answer

    Stored correct answer: A

    This matches our derived answer.

PreviousNext

More from Definite Integration

  • x→0lim​x30∫x2​(sint​)dt​ is equal to :2021 · MCQ
  • If −a∫a​(∣x∣+∣x−2∣)dx=22, (a > 2) and [x] denotes the greatest integer ≤ x, then −a∫a​(x+[x])dx is equal to…2021 · Numerical
  • The value of the integral, 1∫3​[x2−2x−2]dx, where [x] denotes the greatest integer less than or equal to x, is :2021 · MCQ
  • Let f(x) be a differentiable function defined on [0, 2] such that f'(x) = f'(2 − x) for all x ∈ (0, 2), f(0) = 1 and f(2) = e2. Then the value of 0∫2​f(x)dx is :2021 · MCQ
  • Let f be a twice differentiable function defined on R such that f(0) = 1, f'(0) = 2 and f'(x) e 0 for all x ∈ R. If ​f(x)f′(x)​f′(x)f′′(x)​​ = 0, for all x ∈…2021 · MCQ
  • The value of −1∫1​x2e[x3]dx, where [ t ] denotes the greatest integer ≤ t, is :2021 · MCQ
  • If In​=4π​∫2π​​cotnxdx, then :2021 · MCQ
  • The value of −2∫2​∣3x2−3x−6∣dx is ​.2021 · Numerical