What is the integral from 0 to 1 of (x^{19}+19^x) ?

The integral from 0 to 1 of (x^{19}+19^x) is 1/20+(18)/(ln(19))

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integral from 0 to 1 of (x^{19}+19^x)

\int_{0}^{1} (x^{19} + 1 9^{x}) d x

\frac{1}{20} + \frac{18}{ln ( 19 )}

Decimal

6.16321 \dots

Solution steps

\int_{0}^{1} x^{19} + 1 9^{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} x^{19} d x + \int_{0}^{1} 1 9^{x} d x

\int_{0}^{1} x^{19} d x = \frac{1}{20}

\int_{0}^{1} 1 9^{x} d x = \frac{18}{ln ( 19 )}

= \frac{1}{20} + \frac{18}{ln ( 19 )}

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What is the integral from 0 to 1 of (x^{19}+19^x) ?
The integral from 0 to 1 of (x^{19}+19^x) is 1/20+(18)/(ln(19))