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

The integral from-1 to 0 of sqrt(1-x) is (4sqrt(2)-2)/3

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

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

\frac{4 2 - 2}{3}

Decimal Notation

1.21895 \dots

Solution steps

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

Apply u-substitution

: \int_{2}^{1} - u d u

= \int_{2}^{1} - u d u

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

= - \int_{1}^{2} - u d u

Take the constant out:

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

= - (- \int_{1}^{2} u d u)

Apply the Power Rule

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

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

Simplify

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

Compute the boundaries:

\frac{4 2}{3} - \frac{2}{3}

= \frac{4 2}{3} - \frac{2}{3}

Simplify

= \frac{4 2 - 2}{3}

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

\int_{2}^{8} (5 x^{2} + 3 x + 5) d x

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

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

\int_{1}^{3} - \frac{1}{4} x + 5 d x

What is the integral from-1 to 0 of sqrt(1-x) ?
The integral from-1 to 0 of sqrt(1-x) is (4sqrt(2)-2)/3