What is the integral from 0 to 8 of x^e+e^x ?

The integral from 0 to 8 of x^e+e^x is (8^{e+1})/(e+1)+e^8-1

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integral from 0 to 8 of x^e+e^x

\int_{0}^{8} x^{e} + e^{x} d x

\frac{8 ^{e + 1}}{e + 1} + e^{8} - 1

Decimal

3593.15606 \dots

Solution steps

\int_{0}^{8} x^{e} + e^{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}^{8} x^{e} d x + \int_{0}^{8} e^{x} d x

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

\int_{0}^{8} e^{x} d x = e^{8} - 1

= \frac{8 ^{e + 1}}{e + 1} + e^{8} - 1

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What is the integral from 0 to 8 of x^e+e^x ?
The integral from 0 to 8 of x^e+e^x is (8^{e+1})/(e+1)+e^8-1