What is the integral from 0 to 1 of 8xe^{4x}+2e^{4x} ?

The integral from 0 to 1 of 8xe^{4x}+2e^{4x} is 2e^4

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integral from 0 to 1 of 8xe^{4x}+2e^{4x}

\int_{0}^{1} 8 x e^{4 x} + 2 e^{4 x} d x

2 e^{4}

Decimal

109.19630 \dots

Solution steps

\int_{0}^{1} 8 x e^{4 x} + 2 e^{4 x} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{1} 8 x e^{4 x} d x + \int_{0}^{1} 2 e^{4 x} d x

\int_{0}^{1} 8 x e^{4 x} d x = \frac{3 e ^{4} + 1}{2}

\int_{0}^{1} 2 e^{4 x} d x = \frac{e ^{4} - 1}{2}

= \frac{3 e ^{4} + 1}{2} + \frac{e ^{4} - 1}{2}

Simplify

\frac{3 e ^{4} + 1}{2} + \frac{e ^{4} - 1}{2} : 2 e^{4}

= 2 e^{4}

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What is the integral from 0 to 1 of 8xe^{4x}+2e^{4x} ?
The integral from 0 to 1 of 8xe^{4x}+2e^{4x} is 2e^4