The value of the definite integral,intlimits_ | vedclass

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Question

(A) Let $I = \int\limits_0^{100} \frac{x}{e^{x^2}} \,dx$. Substitute $t = x^2$,then $dt = 2x \,dx$,which implies $x \,dx = \frac{dt}{2}$. When $x = 0$

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$\frac{1}{2} (1 - e^{-10000})$

Vedclass · Answer

(A) Let $I = \int\limits_0^{100} \frac{x}{e^{x^2}} \,dx$. Substitute $t = x^2$,then $dt = 2x \,dx$,which implies $x \,dx = \frac{dt}{2}$. When $x = 0$,$t = 0^2 = 0$. When $x = 100$,$t = 100^2 = 10000$. Substituting these into the integral: $I = \int\limits_0^{10000} \frac{1}{e^t} \cdot \frac{dt}{2} = \frac{1}{2} \int\limits_0^{10000} e^{-t} \,dt$. Evaluating the integral: $I = \frac{1}{2} \left[ -e^{-t} \right]_0^{10000} = \frac{1}{2} (-e^{-10000} - (-e^0))$. Since $e^0 = 1$: $I = \frac{1}{2} (1 - e^{-10000})$.

The value of the definite integral,$\int\limits_0^{100} {\frac{x}{{{e^{{x^2}}}}}} \,dx$ is equal to

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