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The value of the integral $\int_{1}^{9} \mathrm{e}^{\sqrt{t}} \mathrm{dt}=$
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Verified Answer
The correct answer is:
$4 \mathrm{e}^{3}$
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 : $x=1 \& x=3$
$\begin{array}{l}
I=\int_{1}^{3} e^{x} \cdot 2 x d x=2 \int_{1}^{3} e^{x} \cdot x d x \\
\Rightarrow I=2\left[\left\{x \cdot e^{x}\right\}_{1}^{3}-\int_{1}^{3} e^{x} d x\right] \\
=2\left[\left\{x e^{x}\right\}_{1}^{3}-\left\{e^{x}\right\}_{1}^{3}\right] \\
\Rightarrow I=2\left[3 e^{3}-e^{1}-e^{3}+e^{1}\right]=\left[2 e^{3}\right] \cdot 2=4 e^{3}
\end{array}$
Put $\mathrm{t}=\mathrm{x}^{2} \Rightarrow \mathrm{dt}=2 \mathrm{x} \cdot \mathrm{dx}$
For limit : $x=1 \& x=3$
$\begin{array}{l}
I=\int_{1}^{3} e^{x} \cdot 2 x d x=2 \int_{1}^{3} e^{x} \cdot x d x \\
\Rightarrow I=2\left[\left\{x \cdot e^{x}\right\}_{1}^{3}-\int_{1}^{3} e^{x} d x\right] \\
=2\left[\left\{x e^{x}\right\}_{1}^{3}-\left\{e^{x}\right\}_{1}^{3}\right] \\
\Rightarrow I=2\left[3 e^{3}-e^{1}-e^{3}+e^{1}\right]=\left[2 e^{3}\right] \cdot 2=4 e^{3}
\end{array}$
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