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

The integral from 0 to 1 of (e^{4x}+2x^2) is (e^4-1)/4+2/3

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integral from 0 to 1 of (e^{4x}+2x^2)

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

\frac{e ^{4} - 1}{4} + \frac{2}{3}

Decimal

14.06620 \dots

Solution steps

\int_{0}^{1} e^{4 x} + 2 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} e^{4 x} d x + \int_{0}^{1} 2 x^{2} d x

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

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

= \frac{e ^{4} - 1}{4} + \frac{2}{3}

\int_{0}^{\infty} e^{- 3 x^{2}} d x

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

\int_{2}^{4} 32 - 8 x d x

\int_{0}^{3} (3 x)^{2} - (x^{2})^{2} d x

\int_{1}^{3} (\frac{5}{x}) d x

What is the integral from 0 to 1 of (e^{4x}+2x^2) ?
The integral from 0 to 1 of (e^{4x}+2x^2) is (e^4-1)/4+2/3