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Area bounded by the curve $y=x e^{x^2}, x$ - axis and the ordinates $x=0, x=a$
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$\frac{e^{a^2}-1}{2}$sq. unit
Required area is $\int_0^a y d x=\int_0^a x e^{x^2} d x$
We put $x^2=t \Rightarrow d x=\frac{d t}{2 x}$ as $x=0 \Rightarrow t=0$ and $x=a \Rightarrow t=a^2$, then it reduces to $\frac{1}{2} \int_0^{a^2} e^t d t=\frac{1}{2}\left[e^t\right]_0^{a^2}=\frac{e^{a^2}-1}{2}$sq. unit.
We put $x^2=t \Rightarrow d x=\frac{d t}{2 x}$ as $x=0 \Rightarrow t=0$ and $x=a \Rightarrow t=a^2$, then it reduces to $\frac{1}{2} \int_0^{a^2} e^t d t=\frac{1}{2}\left[e^t\right]_0^{a^2}=\frac{e^{a^2}-1}{2}$sq. unit.
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