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

Indefinite Integrals question

2019 · 10 Apr · 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. /Indefinite Integrals
  5. /2019 · 10 Apr · Shift 2 · Q28

Indefinite Integrals question

2019 · 10 Apr · Shift 2 · Q28

JEE MainMathematicsIndefinite IntegralsMCQ+4 / −1
If ∫x5e−x2dx=g(x)e−x2+c\int {{x^5}} {e^{ - {x^2}}}dx = g\left( x \right){e^{ - {x^2}}} + c∫x5e−x2dx=g(x)e−x2+c, where c is a constant of integration, then ggg(–1) is equal to :
  1. A
    1
  2. B
    - 1
  3. C
    −52- {5 \over 2}−25​
  4. D
    −12- {1 \over 2}−21​
View written solutionFree

Correct answer: C

  1. We are given
∫x5e−x2 dx=g(x)e−x2+c.\int x^5 e^{-x^2}\,dx = g(x)e^{-x^2}+c.∫x5e−x2dx=g(x)e−x2+c.

We need to find g(−1)g(-1)g(−1).

  1. Differentiate the right-hand form:
ddx(g(x)e−x2)=e−x2(g′(x)−2xg(x)).\frac{d}{dx}\big(g(x)e^{-x^2}\big)=e^{-x^2}\big(g'(x)-2xg(x)\big).dxd​(g(x)e−x2)=e−x2(g′(x)−2xg(x)).

Since this must equal the integrand x5e−x2x^5 e^{-x^2}x5e−x2, we get

g′(x)−2xg(x)=x5.g'(x)-2xg(x)=x^5.g′(x)−2xg(x)=x5.
  1. Instead of solving this differential equation directly, rewrite the integral using substitution. Let
t=x2  ⟹  dt=2x dx.t=x^2 \implies dt=2x\,dx.t=x2⟹dt=2xdx.

Now,

x5dx=x4(xdx)=(x2)2(xdx)=t2dt2.x^5dx=x^4(xdx)=(x^2)^2(xdx)=t^2\frac{dt}{2}.x5dx=x4(xdx)=(x2)2(xdx)=t22dt​.

So the integral becomes

∫x5e−x2dx=12∫t2e−t dt.\int x^5 e^{-x^2}dx=\frac12\int t^2 e^{-t}\,dt.∫x5e−x2dx=21​∫t2e−tdt.
  1. Now evaluate
∫t2e−t dt.\int t^2 e^{-t}\,dt.∫t2e−tdt.

Using the standard result (or integration by parts twice),

∫t2e−t dt=−(t2+2t+2)e−t+C.\int t^2 e^{-t}\,dt=-(t^2+2t+2)e^{-t}+C.∫t2e−tdt=−(t2+2t+2)e−t+C.

Therefore,

∫x5e−x2dx=12[−(t2+2t+2)e−t]+C.\int x^5 e^{-x^2}dx=\frac12\left[-(t^2+2t+2)e^{-t}\right]+C.∫x5e−x2dx=21​[−(t2+2t+2)e−t]+C.

Substitute t=x2t=x^2t=x2:

∫x5e−x2dx=−12(x4+2x2+2)e−x2+C.\int x^5 e^{-x^2}dx=-\frac12(x^4+2x^2+2)e^{-x^2}+C.∫x5e−x2dx=−21​(x4+2x2+2)e−x2+C.

Hence,

g(x)=−12(x4+2x2+2).g(x)=-\frac12(x^4+2x^2+2).g(x)=−21​(x4+2x2+2).
  1. Now compute g(−1)g(-1)g(−1):
g(−1)=−12(1+2+2)=−12⋅5=−52.g(-1)=-\frac12\big(1+2+2\big)=-\frac12\cdot 5=-\frac52.g(−1)=−21​(1+2+2)=−21​⋅5=−25​.
  1. Checking options:
  • A: 111 ❌
  • B: −1-1−1 ❌
  • C: −52-\dfrac52−25​ ✅
  • D: −12-\dfrac12−21​ ❌

Therefore, the correct answer is

−52.\boxed{-\frac52}.−25​​.
PreviousNext

More from Indefinite Integrals

  • Let n ≥ 2 be a natural number and 0<θ<2π​. Then ∫sinn+1θ(sinnθ−sinθ)1/ncosθ​dθ is equal to - (where C is a…2019 · MCQ
  • If ∫ x5.e − 4x3 dx = 481​ e − 4x3 f(x) + C, where C is a constant of inegration, then f(x) is equal to -2019 · MCQ
  • If ∫x41−x2​​ dx = A(x) (1−x2​)m + C, for a suitable chosen integer m and a function A(x), where C is a constant of integration, then (A(x))m equals :2019 · MCQ
  • If ∫2x−1​x+1​dx= f(x) 2x−1​ + C, where C is a constant of integration, then f(x) is equal to :2019 · MCQ
  • The integral ∫x4+x2x3−1​dx is equal to : (Here C is a constant of integration)2019 · MCQ
  • Let a∈(0,2π​) be fixed. If the integral ∫tanx−tanαtanx+tanα​dx = A(x) cos 2 α + B(x) sin 2 α + C, where C is a constant of integration, then the…2019 · MCQ
  • The integral ∫ cos(loge x) dx is equal to : (where C is a constant of integration)2019 · MCQ
  • The integral ∫(2x4+3x2+1)43x13+2x11​dx is equal to : (where C is a constant of integration)2019 · MCQ