If int e^{x^2} cdot x^3 , dx = e^{x^2} f(x) + | vedclass

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(D) Let $I = \int e^{x^2} \cdot x^3 \, dx$. Substitute $t = x^2$,then $dt = 2x \, dx$,which implies $x \, dx = \frac{1}{2} dt$. Substituting these int

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$\frac{x^2-1}{2}$

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(D) Let $I = \int e^{x^2} \cdot x^3 \, dx$. Substitute $t = x^2$,then $dt = 2x \, dx$,which implies $x \, dx = \frac{1}{2} dt$. Substituting these into the integral,we get: $I = \frac{1}{2} \int e^t \cdot t \, dt$. Using integration by parts,$\int u \, dv = uv - \int v \, du$,where $u = t$ and $dv = e^t \, dt$: $I = \frac{1}{2} (t e^t - \int e^t \, dt) = \frac{1}{2} (t e^t - e^t) + c$. $I = \frac{1}{2} e^t (t - 1) + c$. Substituting $t = x^2$ back,we get: $I = \frac{1}{2} e^{x^2} (x^2 - 1) + c$. Comparing this with the given form $e^{x^2} f(x) + c$,we find $f(x) = \frac{x^2 - 1}{2}$. Since $f(1) = \frac{1^2 - 1}{2} = 0$,the condition is satisfied.

If $\int e^{x^2} \cdot x^3 \, dx = e^{x^2} f(x) + c$ and $f(1) = 0$ (where $c$ is a constant of integration),then the value of $f(x)$ is

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