NIMCET 2014 — Mathematics PYQ

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Question

NIMCET 2014 — Mathematics PYQ

NIMCET | Mathematics | 2014

The value of $\int \sqrt{x} e^{\sqrt{x}} dx$ is equal to:

Choose the correct answer:

Explanation

Solution

Concept:

Integration by Parts: $\int f(x) g(x) dx = f(x) \int g(x) dx - \int [f'(x) \int g(x) dx] dx$
Integration by substitution: If we substitute $x = f(t)$ , then $dx = f'(t) dt$ and $\int f(x) dx = \int f[f(t)] f'(t) dt$
$\int e^x dx = e^x + C$

Calculation:

Let $I = \int \sqrt{x} e^{\sqrt{x}} dx$

Substituting $\sqrt{x} = t$ , we get:

\frac{1}{2\sqrt{x}} dx = dt

\Rightarrow dx = 2t dt

\therefore I = \int t \times e^t \times 2t dt

I = 2 \int t^2 e^t dt

Integrating by parts, taking $t^2$ as the first function and $e^t$ as the second function, we get:

\Rightarrow I = 2 \left[ t^2 \int e^t dt - \int \left( \frac{d}{dt} t^2 \int e^t dt \right) dt \right] + C

\Rightarrow I = 2t^2 e^t - 4 \int t e^t dt + C

Integrating $\int t e^t dt$ by parts, we get:

\Rightarrow I = 2t^2 e^t - 4 \left[ t \int e^t dt - \int \left( \frac{d}{dt} t \int e^t dt \right) dt \right] + C

\Rightarrow I = 2t^2 e^t - 4 (t e^t - e^t) + C

\Rightarrow I = (2t^2 - 4t + 4)e^t + C

Back substituting $\sqrt{x} = t$ , we get:

\Rightarrow I = (2x - 4\sqrt{x} + 4)e^{\sqrt{x}} + C

Correct Option: 2. $(2x - 4\sqrt{x} + 4)e^{\sqrt{x}} + C$