What is the integral from 0 to 2 of axe^{ax^2} ?

The integral from 0 to 2 of axe^{ax^2} is 1/2 (e^{4a}-1)

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integral from 0 to 2 of axe^{ax^2}

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

\frac{1}{2} (e^{4 a} - 1)

Solution steps

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

Take the constant out:

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

= a \cdot \int_{0}^{2} x e^{a x^{2}} d x

Apply u-substitution

: \int_{0}^{4 a} \frac{e ^{u}}{2 a} d u

= a \cdot \int_{0}^{4 a} \frac{e ^{u}}{2 a} d u

Take the constant out:

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

= a \frac{1}{2 a} \cdot \int_{0}^{4 a} e^{u} d u

Use the common integral:

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

= a \frac{1}{2 a} [e^{u}]_{0}^{4 a}

Simplify

a \frac{1}{2 a} [e^{u}]_{0}^{4 a} : \frac{1}{2} [e^{u}]_{0}^{4 a}

= \frac{1}{2} [e^{u}]_{0}^{4 a}

Compute the boundaries:

e^{4 a} - 1

= \frac{1}{2} (e^{4 a} - 1)

\int_{3}^{4} (2 x - 1)^{5} d x

\int_{- 5}^{5} x^{2} - 25 d x

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

\int_{3}^{5} 6 x^{2} + 8 d x

\int_{4}^{8} (2 x + 15) e^{0.25 x} d x

What is the integral from 0 to 2 of axe^{ax^2} ?
The integral from 0 to 2 of axe^{ax^2} is 1/2 (e^{4a}-1)