What is the integral from-1 to 0 of 3t^2e^{(-t^3+1)} ?

The integral from-1 to 0 of 3t^2e^{(-t^3+1)} is e^2-e

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integral from-1 to 0 of 3t^2e^{(-t^3+1)}

\int_{- 1}^{0} 3 t^{2} e^{(- t^{3} + 1)} d t

e^{2} - e

Decimal Notation

4.67077 \dots

Solution steps

\int_{- 1}^{0} 3 t^{2} e^{- t^{3} + 1} d t

Take the constant out:

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

= 3 \cdot \int_{- 1}^{0} t^{2} e^{- t^{3} + 1} d t

Apply u-substitution

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

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

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Simplify

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

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

Compute the boundaries:

e^{2} - e

= e^{2} - e

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What is the integral from-1 to 0 of 3t^2e^{(-t^3+1)} ?
The integral from-1 to 0 of 3t^2e^{(-t^3+1)} is e^2-e