Find the area bounded by the curve y = x{e^{{ | vedclass

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(B) The required area $A$ is given by the definite integral: $A = \int_0^a y \, dx = \int_0^a x{e^{{x^2}}} \, dx$ Let $t = x^2$. Then,differentiating

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$\frac{{{e^{{a^2}}} - 1}}{2} 	ext{ sq. unit}$

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(B) The required area $A$ is given by the definite integral: $A = \int_0^a y \, dx = \int_0^a x{e^{{x^2}}} \, dx$ Let $t = x^2$. Then,differentiating both sides with respect to $x$,we get $dt = 2x \, dx$,which implies $x \, dx = \frac{dt}{2}$. Changing the limits of integration: When $x = 0$,$t = 0^2 = 0$. When $x = a$,$t = a^2$. Substituting these into the integral: $A = \int_0^{a^2} {e^t} \cdot \frac{dt}{2} = \frac{1}{2} \int_0^{a^2} {e^t} \, dt$ Evaluating the integral: $A = \frac{1}{2} [e^t]_0^{a^2} = \frac{1}{2} (e^{a^2} - e^0) = \frac{e^{a^2} - 1}{2} 	ext{ sq. unit}$.

Find the area bounded by the curve $y = x{e^{{x^2}}}$,the $x$-axis,and the ordinates $x = 0$ and $x = a$.

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