What is the integral from 2 to 4 of 3xe^{x^2-1} ?

The integral from 2 to 4 of 3xe^{x^2-1} is (3(e^{15}-e^3))/2

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integral from 2 to 4 of 3xe^{x^2-1}

\int_{2}^{4} 3 x e^{x^{2} - 1} d x

\frac{3 ( e ^{15} - e ^{3} )}{2}

Decimal

4903495.93040 \dots

Solution steps

\int_{2}^{4} 3 x e^{x^{2} - 1} d x

Take the constant out:

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

= 3 \cdot \int_{2}^{4} x e^{x^{2} - 1} d x

Apply u-substitution

: \int_{3}^{15} \frac{e ^{u}}{2} d u

= 3 \cdot \int_{3}^{15} \frac{e ^{u}}{2} d u

Take the constant out:

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

= 3 \cdot \frac{1}{2} \cdot \int_{3}^{15} e^{u} d u

Use the common integral:

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

= 3 \cdot \frac{1}{2} [e^{u}]_{3}^{15}

Simplify

3 \cdot \frac{1}{2} [e^{u}]_{3}^{15} : \frac{3}{2} [e^{u}]_{3}^{15}

= \frac{3}{2} [e^{u}]_{3}^{15}

Compute the boundaries:

e^{15} - e^{3}

= \frac{3}{2} (e^{15} - e^{3})

Simplify

= \frac{3 ( e ^{15} - e ^{3} )}{2}

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What is the integral from 2 to 4 of 3xe^{x^2-1} ?
The integral from 2 to 4 of 3xe^{x^2-1} is (3(e^{15}-e^3))/2