What is the integral from 1 to 2 of xe^{2x^2-3} ?

The integral from 1 to 2 of xe^{2x^2-3} is (e^6-1)/(4e)

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integral from 1 to 2 of xe^{2x^2-3}

\int_{1}^{2} x e^{2 x^{2} - 3} d x

\frac{e ^{6} - 1}{4 e}

Decimal

37.01131 \dots

Solution steps

\int_{1}^{2} x e^{2 x^{2} - 3} d x

Apply u-substitution

= \int_{- 1}^{5} \frac{e ^{u}}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int_{- 1}^{5} e^{u} d u

Use the common integral:

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

= \frac{1}{4} [e^{u}]_{- 1}^{5}

Compute the boundaries:

e^{5} - \frac{1}{e}

= \frac{1}{4} (e^{5} - \frac{1}{e})

Simplify

= \frac{e ^{6} - 1}{4 e}

What is the integral from 1 to 2 of xe^{2x^2-3} ?
The integral from 1 to 2 of xe^{2x^2-3} is (e^6-1)/(4e)