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

The integral from 0 to 1 of (2^x+x^2) is 1/(ln(2))+1/3

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

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

\frac{1}{ln ( 2 )} + \frac{1}{3}

Decimal

1.77602 \dots

Solution steps

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

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

\int_{0}^{1} x^{2} d x = \frac{1}{3}

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

\int_{- \infty}^{0} θ e^{θ} d θ

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

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\int_{0}^{1} x^{2} (3 + 4 x^{3})^{3} d x

What is the integral from 0 to 1 of (2^x+x^2) ?
The integral from 0 to 1 of (2^x+x^2) is 1/(ln(2))+1/3