What is the integral from 1 to 2 of 8*e^{4x-3} ?

The integral from 1 to 2 of 8*e^{4x-3} is 2(e^5-e)

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integral from 1 to 2 of 8*e^{4x-3}

\int_{1}^{2} 8 \cdot e^{4 x - 3} d x

2 (e^{5} - e)

Decimal

291.38975 \dots

Solution steps

\int_{1}^{2} 8 e^{4 x - 3} d x

Take the constant out:

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

= 8 \cdot \int_{1}^{2} e^{4 x - 3} d x

Apply u-substitution

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

= 8 \cdot \int_{1}^{5} e^{u} \frac{1}{4} d u

Take the constant out:

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

= 8 \cdot \frac{1}{4} \cdot \int_{1}^{5} e^{u} d u

Use the common integral:

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

= 8 \cdot \frac{1}{4} [e^{u}]_{1}^{5}

Simplify

8 \cdot \frac{1}{4} [e^{u}]_{1}^{5} : 2 [e^{u}]_{1}^{5}

= 2 [e^{u}]_{1}^{5}

Compute the boundaries:

e^{5} - e

= 2 (e^{5} - e)

\int_{0}^{x^{2}} t d t

\int_{- 1}^{2} x^{3} - 2 x^{2} - 5 x + 6 d x

\int_{- 1}^{0} x - 1 d x

\int_{0}^{1} \frac{1}{3 - 2 x ^{2}} d x

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

What is the integral from 1 to 2 of 8*e^{4x-3} ?
The integral from 1 to 2 of 8*e^{4x-3} is 2(e^5-e)