What is the integral from 2 to 3 of 4xe^{x^2-5} ?

The integral from 2 to 3 of 4xe^{x^2-5} is 2(e^4-1/e)

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integral from 2 to 3 of 4xe^{x^2-5}

\int_{2}^{3} 4 x e^{x^{2} - 5} d x

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

Decimal

108.46054 \dots

Solution steps

\int_{2}^{3} 4 x e^{x^{2} - 5} d x

Take the constant out:

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

= 4 \cdot \int_{2}^{3} x e^{x^{2} - 5} d x

Apply u-substitution

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

Take the constant out:

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

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

Use the common integral:

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

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

Simplify

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

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

Compute the boundaries:

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

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

What is the integral from 2 to 3 of 4xe^{x^2-5} ?
The integral from 2 to 3 of 4xe^{x^2-5} is 2(e^4-1/e)