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

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

Popular >

integral from 0 to 1 of sqrt(1-t)

\int_{0}^{1} 1 - t d t

\frac{2}{3}

Decimal Notation

0.66666 \dots

Solution steps

\int_{0}^{1} 1 - t d t

Apply u-substitution

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

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

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

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

Take the constant out:

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

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

Apply the Power Rule

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

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

Simplify

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

Compute the boundaries:

\frac{2}{3}

= \frac{2}{3}

\int_{0}^{3} (9 - x^{2}) d x

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

\int_{0}^{1} (5 x^{2} - 6 x + 6) d x

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

\int_{0}^{1} cos (πn x) d x

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