What is the integral from 0 to 5 of e^{sqrt(3x+1)} ?

The integral from 0 to 5 of e^{sqrt(3x+1)} is 2e^4

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integral from 0 to 5 of e^{sqrt(3x+1)}

\int_{0}^{5} e^{3 x + 1} d x

2 e^{4}

Decimal

109.19630 \dots

Solution steps

\int_{0}^{5} e^{3 x + 1} d x

Apply u-substitution

: \int_{1}^{4} \frac{2 e ^{u} u}{3} d u

= \int_{1}^{4} \frac{2 e ^{u} u}{3} d u

Take the constant out:

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

= \frac{2}{3} \cdot \int_{1}^{4} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{1}^{4}

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

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

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

Compute the boundaries:

3 e^{4}

= \frac{2}{3} \cdot 3 e^{4}

Simplify

= 2 e^{4}

\int_{0}^{\infty} 5 e^{- 5 t} d t

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\int_{0}^{4} \frac{1}{9 x ^{2} + 25} d x

What is the integral from 0 to 5 of e^{sqrt(3x+1)} ?
The integral from 0 to 5 of e^{sqrt(3x+1)} is 2e^4