What is the integral from 0 to 1 of (x^{16}+16^x) ?

The integral from 0 to 1 of (x^{16}+16^x) is 1/17+(15)/(4ln(2))

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integral from 0 to 1 of (x^{16}+16^x)

\int_{0}^{1} (x^{16} + 1 6^{x}) d x

\frac{1}{17} + \frac{15}{4 ln ( 2 )}

Decimal

5.46892 \dots

Solution steps

\int_{0}^{1} x^{16} + 1 6^{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^{16} d x + \int_{0}^{1} 1 6^{x} d x

\int_{0}^{1} x^{16} d x = \frac{1}{17}

\int_{0}^{1} 1 6^{x} d x = \frac{15}{4 ln ( 2 )}

= \frac{1}{17} + \frac{15}{4 ln ( 2 )}

What is the integral from 0 to 1 of (x^{16}+16^x) ?
The integral from 0 to 1 of (x^{16}+16^x) is 1/17+(15)/(4ln(2))