What is the integral from-1 to 2 of 3x^2e^{x^3+5} ?

The integral from-1 to 2 of 3x^2e^{x^3+5} is e^{13}-e^4

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integral from-1 to 2 of 3x^2e^{x^3+5}

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

e^{13} - e^{4}

Decimal Notation

442358.79385 \dots

Solution steps

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

Apply u-substitution

: \int_{4}^{13} \frac{e ^{u}}{3} d u

= 3 \cdot \int_{4}^{13} \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 \cdot \frac{1}{3} \cdot \int_{4}^{13} e^{u} d u

Use the common integral:

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

= 3 \cdot \frac{1}{3} [e^{u}]_{4}^{13}

Simplify

3 \cdot \frac{1}{3} [e^{u}]_{4}^{13} : [e^{u}]_{4}^{13}

= [e^{u}]_{4}^{13}

Compute the boundaries:

e^{13} - e^{4}

= e^{13} - e^{4}

\int_{2}^{6} \frac{1}{6 - x} d x

\int_{0}^{1} \frac{1}{9 - x ^{2}} d x

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

\int_{0}^{1} x^{4} \cdot e^{- x} d x

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

What is the integral from-1 to 2 of 3x^2e^{x^3+5} ?
The integral from-1 to 2 of 3x^2e^{x^3+5} is e^{13}-e^4