What is the integral from 0 to 1 of f(sqrt(1-x)) ?

The integral from 0 to 1 of f(sqrt(1-x)) is (2f)/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 1 of f(sqrt(1-x))

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

\frac{2 f}{3}

Solution steps

\int_{0}^{1} f 1 - x d x

Take the constant out:

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

= f \cdot \int_{0}^{1} 1 - x d x

Apply u-substitution

: \int_{1}^{0} - u d u

= f \cdot \int_{1}^{0} - u d u

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

= f (- \int_{0}^{1} - u d u)

Take the constant out:

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

= f (- (- \int_{0}^{1} u d u))

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{1}

= f (- (- [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{1}))

Simplify

= f [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{1}

Compute the boundaries:

\frac{2}{3}

= f \frac{2}{3}

Simplify

= \frac{2 f}{3}

\int_{- 2}^{2} (11 - ∣ x ∣^{3}) d x

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

\int_{1}^{4} \frac{3}{1 + 4 x ^{2}} d x

\int_{0}^{2} 16 - 4 x^{2} d x

\int_{- \infty}^{1} e^{- x} d x

What is the integral from 0 to 1 of f(sqrt(1-x)) ?
The integral from 0 to 1 of f(sqrt(1-x)) is (2f)/3