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NDA 2010
Exam
NDA
Year
2010
Subject
Mathematics
Chapter
Indefinite Integration

NDA 2010 · Mathematics · Indefinite Integration

What is ∫ x e^ x d x equal to? where c is a constant of integration.

What is ∫xexdx\int \sqrt{x} e^{\sqrt{x}} d x equal to? where cc is a constant of integration.

Options

  1. A2ex(x−2x+2)+c2 e^{\sqrt{x}}(x-2 \sqrt{x}+2)+c
  2. B2ex(x+2x+2)+c2 e^{\sqrt{x}}(x+2 \sqrt{x}+2)+c
  3. C2ex(x+2x−2)+c2 e^{\sqrt{x}}(x+2 \sqrt{x}-2)+c
  4. D2ex(x−2x−2)+c2 e^{\sqrt{x}}(x-2 \sqrt{x}-2)+c

Correct answer

A. 2ex(x−2x+2)+c2 e^{\sqrt{x}}(x-2 \sqrt{x}+2)+c

Step-by-step solution

∫xexdx\int \sqrt{x} e^{\sqrt{x}} d x Put x=u\sqrt{x}=u12x=du∴dx=2xdu=∫u2eudu×2=2[u2∫eudu−∫2ueudu]=2[u2⋅eu−2{eu⋅u−eu}]=2u2eu−4ueu+4eu=2xex−4xex+4ex=2ex(x−2x+2)\begin{aligned} & \frac{1}{2 \sqrt{x}}=d u \\ & \therefore d x=2 \sqrt{x} d u \\ &= \int u^2 e^u d u \times 2 \\ &= 2\left[u^2 \int e^u d u-\int 2 u e^u d u\right] \\ &= 2\left[u^2 \cdot e^u-2\left\{e^u \cdot u-e^u\right\}\right] \\ &= 2 u^2 e^u-4 u e^u+4 e^u \\ &= 2 x e^{\sqrt{x}}-4 \sqrt{x} e^{\sqrt{x}}+4 e^{\sqrt{x}} \\ &= 2 e^{\sqrt{x}}(x-2 \sqrt{x}+2)\end{aligned}$$
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