What is the integral from 0 to b of (2^x+x^2) ?

The integral from 0 to b of (2^x+x^2) is (2^b-1}{ln(2)}+\frac{b^3)/3

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integral from 0 to b of (2^x+x^2)

\int_{0}^{b} (2^{x} + x^{2}) d x

\frac{2 ^{b} - 1}{ln ( 2 )} + \frac{b ^{3}}{3}

Solution steps

\int_{0}^{b} 2^{x} + x^{2} 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}^{b} 2^{x} d x + \int_{0}^{b} x^{2} d x

\int_{0}^{b} 2^{x} d x = \frac{2 ^{b} - 1}{ln ( 2 )}

\int_{0}^{b} x^{2} d x = \frac{b ^{3}}{3}

= \frac{2 ^{b} - 1}{ln ( 2 )} + \frac{b ^{3}}{3}

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What is the integral from 0 to b of (2^x+x^2) ?
The integral from 0 to b of (2^x+x^2) is (2^b-1}{ln(2)}+\frac{b^3)/3