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

The integral from 0 to 1 of (x^3+1)e^{2x} is (5e^2-1)/8

Popular >

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

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

\frac{5 e ^{2} - 1}{8}

Decimal

4.49316 \dots

Solution steps

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

Expand

(x^{3} + 1) e^{2 x} : e^{2 x} x^{3} + e^{2 x}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int_{0}^{1} e^{2 x} x^{3} d x = \frac{e ^{2} + 3}{8}

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

= \frac{e ^{2} + 3}{8} + \frac{e ^{2} - 1}{2}

Simplify

\frac{e ^{2} + 3}{8} + \frac{e ^{2} - 1}{2} : \frac{5 e ^{2} - 1}{8}

= \frac{5 e ^{2} - 1}{8}

\int_{c o s (x)}^{1} e^{- t^{2}} d t

\int_{0}^{1} 4 x^{2} d x

\int_{1}^{e^{2}} \frac{ln ( x )}{x} d x

\int_{0}^{1} 3 cos (\frac{π t}{2}) d t

\int_{1}^{7} \frac{e ^{\frac{1}{x}}}{x ^{2}} d x

What is the integral from 0 to 1 of (x^3+1)e^{2x} ?
The integral from 0 to 1 of (x^3+1)e^{2x} is (5e^2-1)/8