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

The integral from 0 to 1 of (3-4x)e^{2x} is (e^2-5)/2

Popular >

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

\int_{0}^{1} (3 - 4 x) e^{2 x} d x

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

Decimal

1.19452 \dots

Solution steps

\int_{0}^{1} (3 - 4 x) e^{2 x} d x

Apply Integration By Parts

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

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

\int - 2 e^{2 x} d x = - e^{2 x}

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

Simplify

[\frac{1}{2} e^{2 x} (3 - 4 x) - (- e^{2 x})]_{0}^{1} : [\frac{1}{2} (- 4 e^{2 x} x + 5 e^{2 x})]_{0}^{1}

= [\frac{1}{2} (- 4 e^{2 x} x + 5 e^{2 x})]_{0}^{1}

Compute the boundaries:

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

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

Simplify

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

y^{^{''}} + \frac{200}{8} \cdot y = 30 sin (2 x), y (0) = 2, y^{^{'}} (0) = 0

\frac{d}{d x} (- \frac{c}{x ^{2}})

\frac{2 x + 1}{x + 2}

3 t^{4} e^{- 2 t}

x^{5}, e^{3} x^{2}

What is the integral from 0 to 1 of (3-4x)e^{2x} ?
The integral from 0 to 1 of (3-4x)e^{2x} is (e^2-5)/2