What is the integral from 0 to 1 of x^2e^{x^3+0} ?

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

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integral from 0 to 1 of x^2e^{x^3+0}

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

\frac{e - 1}{3}

Decimal Notation

0.57276 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{1} \frac{e ^{u}}{3} d u

= \int_{0}^{1} \frac{e ^{u}}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int_{0}^{1} e^{u} d u

Use the common integral:

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

= \frac{1}{3} [e^{u}]_{0}^{1}

Compute the boundaries:

e - 1

= \frac{1}{3} (e - 1)

Simplify

= \frac{e - 1}{3}

y^{^{''}} + y = x

f (x) = 3 x^{2} + 3 x, at x = - 2

\int ∣ 5 x ∣ d x

f (x) = \frac{53}{0.01 x ^{2} + 1}

\int \frac{e ^{- 4 x}}{x ^{2}} d x

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