If f(x) = int_a^x {t^3 e^t , dt},then frac{d} | vedclass

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(B) According to the Leibniz Integral Rule,if $f(x) = \int_{g(x)}^{h(x)} F(t, x) \, dt$,then $\frac{d}{dx} f(x) = F(h(x), x) \cdot h'(x) - F(g(x), x)

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$x^3 e^x$

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(B) According to the Leibniz Integral Rule,if $f(x) = \int_{g(x)}^{h(x)} F(t, x) \, dt$,then $\frac{d}{dx} f(x) = F(h(x), x) \cdot h'(x) - F(g(x), x) \cdot g'(x)$.Given $f(x) = \int_a^x t^3 e^t \, dt$.Here,the integrand is $F(t) = t^3 e^t$,the upper limit is $h(x) = x$,and the lower limit is $g(x) = a$.Applying the rule:$\frac{d}{dx} f(x) = (x^3 e^x) \cdot \frac{d}{dx}(x) - (a^3 e^a) \cdot \frac{d}{dx}(a)$.Since $a$ is a constant,$\frac{d}{dx}(a) = 0$.Therefore,$\frac{d}{dx} f(x) = x^3 e^x \cdot 1 - 0 = x^3 e^x$.

If $f(x) = \int_a^x {t^3 e^t \, dt}$,then $\frac{d}{dx} f(x) = $

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