What is the integral from 0 to 4 of (2^x-x^e) ?

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

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integral from 0 to 4 of (2^x-x^e)

\int_{0}^{4} (2^{x} - x^{e}) d x

\frac{15}{ln ( 2 )} - \frac{4 ^{e + 1}}{e + 1}

Decimal

- 24.94889 \dots

Solution steps

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

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

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

= \frac{15}{ln ( 2 )} - \frac{4 ^{e + 1}}{e + 1}

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