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

The integral from 0 to 1 of (x^2+1)e^{2x} is (3e^2-3)/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

\frac{3 e ^{2} - 3}{4}

Decimal

4.79179 \dots

Solution steps

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

Apply Integration By Parts

= [\frac{1}{2} e^{2 x} (x^{2} + 1) - \int e^{2 x} x d x]_{0}^{1}

\int e^{2 x} x d x = \frac{1}{4} (e^{2 x} \cdot 2 x - e^{2 x})

= [\frac{1}{2} e^{2 x} (x^{2} + 1) - \frac{1}{4} (e^{2 x} \cdot 2 x - e^{2 x})]_{0}^{1}

Simplify

[\frac{1}{2} e^{2 x} (x^{2} + 1) - \frac{1}{4} (e^{2 x} \cdot 2 x - e^{2 x})]_{0}^{1} : [\frac{1}{4} (2 e^{2 x} x^{2} - 2 e^{2 x} x + 3 e^{2 x})]_{0}^{1}

= [\frac{1}{4} (2 e^{2 x} x^{2} - 2 e^{2 x} x + 3 e^{2 x})]_{0}^{1}

Compute the boundaries:

\frac{3 e ^{2}}{4} - \frac{3}{4}

= \frac{3 e ^{2}}{4} - \frac{3}{4}

Simplify

= \frac{3 e ^{2} - 3}{4}

What is the integral from 0 to 1 of (x^2+1)e^{2x} ?
The integral from 0 to 1 of (x^2+1)e^{2x} is (3e^2-3)/4