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JEE Main 201910 AprilEvening Shift
Exam
JEE Main
Year
2019
Date
10 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Indefinite Integration

JEE Main 2019 · 10 April · Evening Shift · Mathematics · Indefinite Integration

If ∫ x ^ 5 e ^ - x ^ 2 dx = g(x) e ^ - x ^ 2 + c , where c is a...

If ∫x5e−x2dx=g(x)e−x2+c∫ {x}^{5}{e}^{- {x}^{2}}dx = g\left(x\right){e}^{- {x}^{2}} + c, where cc is a constant of integration, then g(−1)g\left(- 1\right) is equal to

Options

  1. A−52- \dfrac{5}{2}
  2. B−1- 1
  3. C11
  4. D−12- \dfrac{1}{2}

Correct answer

A. −52- \dfrac{5}{2}

Step-by-step solution

Let I=∫x5e−x2dxI = ∫ {x}^{5}{e}^{- {x}^{2}}dx

Let, −x2=t⇒xdx=−12dt- {x}^{2} = t ⇒ xdx = - \dfrac{1}{2}dt

∴I=−12∫t2etdt∴ I = - \dfrac{1}{2} ∫ {t}^{2}{e}^{t}dt

By using integrating by parts, i.e. ∫(u⋅v)dx=u∫vdx−∫[ddxu∫vdx]dx+c,∫ \left(u \cdot v\right)dx = u ∫ vdx - ∫ \left[\dfrac{d}{dx}u ∫ vdx\right]dx + c, we get

I=−12[t2∫etdt−∫(2t∫etdt)dt]+cI = - \dfrac{1}{2}\left[{t}^{2} ∫ {e}^{t}dt - ∫ \left(2t ∫ {e}^{t}dt\right)dt\right] + c

⇒I=−12[t2et−2∫tetdt]+c⇒ I = - \dfrac{1}{2}\left[{t}^{2}{e}^{t} - 2 ∫ t{e}^{t}dt\right] + c

Again, applying integrating by parts, we get

⇒I=−12[t2et−2(t∫etdt−∫1⋅etdt)]+c⇒ I = - \dfrac{1}{2}\left[{t}^{2}{e}^{t} - 2\left(t ∫ {e}^{t}dt - ∫ 1 \cdot {e}^{t}dt\right)\right] + c

⇒I=−12[t2et−2(tet−et)]+c⇒ I = - \dfrac{1}{2}\left[{t}^{2}{e}^{t} - 2\left(t{e}^{t} - {e}^{t}\right)\right] + c

⇒I=−12[t2et−2tet+2et]+c⇒ I = - \dfrac{1}{2}\left[{t}^{2}{e}^{t} - 2t{e}^{t} + 2{e}^{t}\right] + c

⇒I=−et2[t2−2t+2]+c⇒ I = - \dfrac{{e}^{t}}{2}\left[{t}^{2} - 2t + 2\right] + c

⇒I=−e−x22[x4+2x2+2]+c⇒ I = - \dfrac{{e}^{- {x}^{2}}}{2}\left[{x}^{4} + 2{x}^{2} + 2\right] + c

Given, I=∫x5e−x2dx=g(x)e−x2+cI = ∫ {x}^{5}{e}^{- {x}^{2}}dx = g\left(x\right){e}^{- {x}^{2}} + c

⇒g(x)=−12(x4+2x2+2)⇒ g\left(x\right) = - \dfrac{1}{2}\left({x}^{4} + 2{x}^{2} + 2\right)

⇒g(−1)=−12((−1)4+2(−1)2+2)⇒ g\left(- 1\right) = - \dfrac{1}{2}\left({\left(- 1\right)}^{4} + 2{\left(- 1\right)}^{2} + 2\right)

⇒g(−1)=−52.⇒ g\left(- 1\right) = - \dfrac{5}{2} .

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