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

The integral from 0 to 1 of e^x(4-e^x) is (-e^2+8e-7)/2

Popular >

integral from 0 to 1 of e^x(4-e^x)

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

\frac{- e ^{2} + 8 e - 7}{2}

Decimal

3.67859 \dots

Solution steps

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

Apply u-substitution

: \int_{3}^{4 - e} - u d u

= \int_{3}^{4 - e} - u d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{4 - e}^{3} - u d u

Take the constant out:

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

= - (- \int_{4 - e}^{3} u d u)

Apply the Power Rule

: [\frac{u ^{2}}{2}]_{4 - e}^{3}

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

Simplify

= [\frac{u ^{2}}{2}]_{4 - e}^{3}

Compute the boundaries:

\frac{9}{2} - \frac{( 4 - e ) ^{2}}{2}

= \frac{9}{2} - \frac{( 4 - e ) ^{2}}{2}

Simplify

= \frac{- e ^{2} + 8 e - 7}{2}

n = 1 \sum \infty - \frac{1}{2}

\int \frac{( x ^{2} )}{25 - x ^{2}} d x

\int (- 9.8 x + 30) d x

csc (7 x) - csc (5 x) + sec (12 x)

y ln (x) (\frac{d y}{d x}) = (\frac{y + 1}{x})^{2}

What is the integral from 0 to 1 of e^x(4-e^x) ?
The integral from 0 to 1 of e^x(4-e^x) is (-e^2+8e-7)/2