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

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

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

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

\frac{3 ( e ^{2} - 1 )}{5}

Decimal

3.83343 \dots

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int_{2}^{0} - \frac{e ^{u}}{5} d u

= 3 \cdot \int_{2}^{0} - \frac{e ^{u}}{5} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= 3 (- \int_{0}^{2} - \frac{e ^{u}}{5} d u)

Take the constant out:

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

= 3 (- (- \frac{1}{5} \cdot \int_{0}^{2} e^{u} d u))

Use the common integral:

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

= 3 (- (- \frac{1}{5} [e^{u}]_{0}^{2}))

Simplify

3 (- (- \frac{1}{5} [e^{u}]_{0}^{2})) : \frac{3}{5} [e^{u}]_{0}^{2}

= \frac{3}{5} [e^{u}]_{0}^{2}

Compute the boundaries:

e^{2} - 1

= \frac{3}{5} (e^{2} - 1)

Simplify

= \frac{3 ( e ^{2} - 1 )}{5}

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