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

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

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

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

\frac{3 e ^{2}}{4} + \frac{1}{4 e ^{2}}

Decimal

5.57562 \dots

Solution steps

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

Apply Integration By Parts

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

\int \frac{1}{2} e^{2 x} d x = \frac{1}{4} e^{2 x}

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

Simplify

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

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

Compute the boundaries:

\frac{3 e ^{2}}{4} + \frac{1}{4 e ^{2}}

= \frac{3 e ^{2}}{4} + \frac{1}{4 e ^{2}}

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