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

The integral from-1 to 1 of sqrt(1-t) is (4sqrt(2))/3

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

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

\frac{4 2}{3}

Decimal

1.88561 \dots

Solution steps

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

Apply u-substitution

: \int_{2}^{0} - u d u

= \int_{2}^{0} - u d u

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

= - \int_{0}^{2} - u d u

Take the constant out:

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

= - (- \int_{0}^{2} u d u)

Apply the Power Rule

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

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

Simplify

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

Compute the boundaries:

\frac{4 2}{3}

= \frac{4 2}{3}

\int_{0}^{1 - z - y} z d z

\int_{1}^{3} - \frac{1}{t ^{3}} - 3 d t

\int_{0}^{1} \frac{11}{1 + p ^{2}} d p

\int_{1}^{\infty} e^{- 2 x} x d x

\int_{0}^{1} r ln (r) d r

What is the integral from-1 to 1 of sqrt(1-t) ?
The integral from-1 to 1 of sqrt(1-t) is (4sqrt(2))/3