Q68 / 120 · VITEEE 2006 · Mathematics

The value of the integral $\int_{1}^{9} \mathrm{e}^{\sqrt{t}} \mathrm{dt}=$

  • A
    $\mathrm{e}^{3}$
  • Correct answer
    B
    $4 \mathrm{e}^{3}$
  • C
    $4\left(\mathrm{e}^{3}-\mathrm{e}\right)$
  • D
    $4 \mathrm{e}^{3}-2 \mathrm{e}$

Solution

Let $\mathrm{I}=\int_{1}^{9} \mathrm{e}^{\sqrt{\mathrm{t}}} \mathrm{dt}$ Put $\mathrm{t}=\mathrm{x}^{2} \Rightarrow \mathrm{dt}=2 \mathrm{x} \cdot \mathrm{dx}$ For limit : $x=1 \& x=3$ $\begin{array}{l} I=\int_{1}^{3} e^{x} \cdot 2 x d x=2 \int_{1}^{3} e^{x} \cdot x d x \\ \Rightarrow I=2\left[\left\{x \cdot e^{x}\right\}_{1}^{3}-\int_{1}^{3} e^{x} d x\right] \\ =2\left[\left\{x e^{x}\right\}_{1}^{3}-\left\{e^{x}\right\}_{1}^{3}\right] \\ \Rightarrow I=2\left[3 e^{3}-e^{1}-e^{3}+e^{1}\right]=\left[2 e^{3}\right] \cdot 2=4 e^{3} \end{array}$

Hence, the correct answer is option: 'B'.

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