What is the integral from 0 to 1 of 2t^4e^{t^5} ?

The integral from 0 to 1 of 2t^4e^{t^5} is (2(e-1))/5

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integral from 0 to 1 of 2t^4e^{t^5}

\int_{0}^{1} 2 t^{4} e^{t^{5}} d t

\frac{2 ( e - 1 )}{5}

Decimal Notation

0.68731 \dots

Solution steps

\int_{0}^{1} 2 t^{4} e^{t^{5}} d t

Take the constant out:

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

= 2 \cdot \int_{0}^{1} t^{4} e^{t^{5}} d t

Apply u-substitution

: \int_{0}^{1} \frac{e ^{u^{\frac{5}{4}}} u ^{\frac{1}{4}}}{4} d u

= 2 \cdot \int_{0}^{1} \frac{e ^{u^{\frac{5}{4}}} u ^{\frac{1}{4}}}{4} d u

Take the constant out:

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

= 2 \cdot \frac{1}{4} \cdot \int_{0}^{1} e^{u^{\frac{5}{4}}} u^{\frac{1}{4}} d u

Apply u-substitution

: \int_{0}^{1} \frac{4 e ^{v}}{5} d v

= 2 \cdot \frac{1}{4} \cdot \int_{0}^{1} \frac{4 e ^{v}}{5} d v

Take the constant out:

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

= 2 \cdot \frac{1}{4} \cdot \frac{4}{5} \cdot \int_{0}^{1} e^{v} d v

Use the common integral:

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

= 2 \cdot \frac{1}{4} \cdot \frac{4}{5} [e^{v}]_{0}^{1}

Simplify

2 \cdot \frac{1}{4} \cdot \frac{4}{5} [e^{v}]_{0}^{1} : \frac{2}{5} [e^{v}]_{0}^{1}

= \frac{2}{5} [e^{v}]_{0}^{1}

Compute the boundaries:

e - 1

= \frac{2}{5} (e - 1)

Simplify

= \frac{2 ( e - 1 )}{5}

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What is the integral from 0 to 1 of 2t^4e^{t^5} ?
The integral from 0 to 1 of 2t^4e^{t^5} is (2(e-1))/5