What is the integral from 0 to x^2 of e^xf(t) ?

The integral from 0 to x^2 of e^xf(t) is (x^4fe^x)/2

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integral from 0 to x^2 of e^xf(t)

\int_{0}^{x^{2}} e^{x} f (t) d t

\frac{x ^{4} f e ^{x}}{2}

Solution steps

\int_{0}^{x^{2}} e^{x} f t d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= f e^{x} \cdot \int_{0}^{x^{2}} t d t

Apply the Power Rule

: [\frac{t ^{2}}{2}]_{0}^{x^{2}}

= f e^{x} [\frac{t ^{2}}{2}]_{0}^{x^{2}}

Compute the boundaries:

\frac{x ^{4}}{2}

= f e^{x} \frac{x ^{4}}{2}

Simplify

= \frac{x ^{4} f e ^{x}}{2}

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What is the integral from 0 to x^2 of e^xf(t) ?
The integral from 0 to x^2 of e^xf(t) is (x^4fe^x)/2