What is the integral from 1 to 3 of 4e^{2x-5} ?

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

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integral from 1 to 3 of 4e^{2x-5}

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

2 (e - \frac{1}{e ^{3}})

Decimal

5.33698 \dots

Solution steps

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

Take the constant out:

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

= 4 \cdot \int_{1}^{3} e^{2 x - 5} d x

Apply u-substitution

: \int_{- 3}^{1} e^{u} \frac{1}{2} d u

= 4 \cdot \int_{- 3}^{1} e^{u} \frac{1}{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_{- 3}^{1} e^{u} d u

Use the common integral:

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

= 4 \cdot \frac{1}{2} [e^{u}]_{- 3}^{1}

Simplify

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

= 2 [e^{u}]_{- 3}^{1}

Compute the boundaries:

e - \frac{1}{e ^{3}}

= 2 (e - \frac{1}{e ^{3}})

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