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

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Question

(A) Given $\int e^{x^2} \cdot x^3 \, dx = e^{x^2} f(x) + C$. Let $I = \int e^{x^2} \cdot x^2 \cdot x \, dx$. Substitute $t = x^2$,then $dt = 2x \, dx$

Vedclass · Accepted Answer

$\frac{3}{2}$

Vedclass · Answer

(A) Given $\int e^{x^2} \cdot x^3 \, dx = e^{x^2} f(x) + C$. Let $I = \int e^{x^2} \cdot x^2 \cdot x \, dx$. Substitute $t = x^2$,then $dt = 2x \, dx$,so $x \, dx = \frac{1}{2} dt$. $I = \frac{1}{2} \int t e^t \, dt$. Using integration by parts: $\int u \, dv = uv - \int v \, du$. Let $u = t$ and $dv = e^t \, dt$,then $du = dt$ and $v = e^t$. $I = \frac{1}{2} [t e^t - \int e^t \, dt] = \frac{1}{2} [t e^t - e^t] + C = \frac{1}{2} e^t (t - 1) + C$. Substituting $t = x^2$ back: $I = \frac{1}{2} e^{x^2} (x^2 - 1) + C$. Comparing with $e^{x^2} f(x) + C$,we get $f(x) = \frac{1}{2} (x^2 - 1)$. Check $f(1) = \frac{1}{2} (1^2 - 1) = 0$,which is consistent. Therefore,$f(2) = \frac{1}{2} (2^2 - 1) = \frac{1}{2} (4 - 1) = \frac{3}{2}$.

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

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