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

The integral from 0 to 4 of (e^x)^2 is (e^8-1)/2

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

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

\frac{e ^{8} - 1}{2}

Decimal Notation

1489.97899 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{e^{4}} u d u

= \int_{1}^{e^{4}} u d u

Apply the Power Rule

: [\frac{u ^{2}}{2}]_{1}^{e^{4}}

= [\frac{u ^{2}}{2}]_{1}^{e^{4}}

Compute the boundaries:

\frac{e ^{8}}{2} - \frac{1}{2}

= \frac{e ^{8}}{2} - \frac{1}{2}

Simplify

= \frac{e ^{8} - 1}{2}

\frac{\partial}{\partial z} (3 y z^{2})

\int_{1}^{10} 5 x - 1 d x

\frac{d}{d x} (x^{- 3} y^{- 4})

n \to \infty lim (n^{0})

f (x) = 3^{(x^{5})}

What is the integral from 0 to 4 of (e^x)^2 ?
The integral from 0 to 4 of (e^x)^2 is (e^8-1)/2