What is the integral from 0 to 1 of (2x^e+9e^x) ?

The integral from 0 to 1 of (2x^e+9e^x) is 2/(e+1)+9(e-1)

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integral from 0 to 1 of (2x^e+9e^x)

\int_{0}^{1} (2 x^{e} + 9 e^{x}) d x

\frac{2}{e + 1} + 9 (e - 1)

Decimal

16.00241 \dots

Solution steps

\int_{0}^{1} 2 x^{e} + 9 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}^{1} 2 x^{e} d x + \int_{0}^{1} 9 e^{x} d x

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

\int_{0}^{1} 9 e^{x} d x = 9 (e - 1)

= \frac{2}{e + 1} + 9 (e - 1)

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What is the integral from 0 to 1 of (2x^e+9e^x) ?
The integral from 0 to 1 of (2x^e+9e^x) is 2/(e+1)+9(e-1)