What is the integral from 0 to 4 of xe^{x^2} ?

The integral from 0 to 4 of xe^{x^2} is (e^{16}-1)/2

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integral from 0 to 4 of xe^{x^2}

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

\frac{e ^{16} - 1}{2}

Decimal Notation

4443054.76025 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{16} \frac{e ^{u}}{2} d u

= \int_{0}^{16} \frac{e ^{u}}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{0}^{16} e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= \frac{1}{2} [e^{u}]_{0}^{16}

Compute the boundaries:

e^{16} - 1

= \frac{1}{2} (e^{16} - 1)

Simplify

= \frac{e ^{16} - 1}{2}

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

\frac{d}{d x} (sin (2) (x))

- \frac{y}{x + e ^{y}}

\int_{0}^{3} \frac{6 t}{( t - 4 ) ^{2}} d t

\int \frac{2 x e ^{x}}{( x + 1 ) ^{2}} d x

What is the integral from 0 to 4 of xe^{x^2} ?
The integral from 0 to 4 of xe^{x^2} is (e^{16}-1)/2