What is the integral from a to b of ((ax)^3+e^{-bx}) ?

The integral from a to b of ((ax)^3+e^{-bx}) is (a^3(b^4-a^4))/4-(e^{-b^2}-e^{-ab})/b

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from a to b of ((ax)^3+e^{-bx})

\int_{a}^{b} ((a x)^{3} + e^{- b x}) d x

\frac{a ^{3} ( b ^{4} - a ^{4} )}{4} - \frac{e ^{- b^{2}} - e ^{- ab}}{b}

Solution steps

\int_{a}^{b} (a x)^{3} + e^{- b x} d x

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

= \int_{a}^{b} a^{3} x^{3} + e^{- b 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_{a}^{b} a^{3} x^{3} d x + \int_{a}^{b} e^{- b x} d x

\int_{a}^{b} a^{3} x^{3} d x = \frac{a ^{3} ( b ^{4} - a ^{4} )}{4}

\int_{a}^{b} e^{- b x} d x = - \frac{e ^{- b^{2}} - e ^{- ab}}{b}

= \frac{a ^{3} ( b ^{4} - a ^{4} )}{4} - \frac{e ^{- b^{2}} - e ^{- ab}}{b}

\int_{0}^{1} c (1 - x^{3}) d x

\int_{0}^{t} e^{x} cos (x) d x

\int_{2}^{3} \frac{4}{x ^{3}} d x

\int_{0}^{3} (e^{\frac{x}{3}} - 3) d x

\int_{0}^{1} (8 ln (x)) d x

What is the integral from a to b of ((ax)^3+e^{-bx}) ?
The integral from a to b of ((ax)^3+e^{-bx}) is (a^3(b^4-a^4))/4-(e^{-b^2}-e^{-ab})/b