What is the integral from 0 to 1 of-sqrt(1-x^2) ?

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

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integral from 0 to 1 of-sqrt(1-x^2)

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

- \frac{π}{4}

Decimal

- 0.78539 \dots

Solution steps

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

Take the constant out:

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

= - \int_{0}^{1} 1 - x^{2} d x

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{2}} cos^{2} (u) d u

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

Rewrite using trig identities

= - \int_{0}^{\frac{π}{2}} \frac{1 + cos ( 2 u )}{2} d u

Take the constant out:

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

= - \frac{1}{2} \cdot \int_{0}^{\frac{π}{2}} 1 + cos (2 u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \frac{1}{2} (\int_{0}^{\frac{π}{2}} 1 d u + \int_{0}^{\frac{π}{2}} cos (2 u) d u)

\int_{0}^{\frac{π}{2}} 1 d u = \frac{π}{2}

\int_{0}^{\frac{π}{2}} cos (2 u) d u = 0

= - \frac{1}{2} (\frac{π}{2} + 0)

Simplify

- \frac{1}{2} (\frac{π}{2} + 0) : - \frac{π}{4}

= - \frac{π}{4}

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

\int_{0}^{3} x^{2} e^{x} d x

\int_{\frac{π}{6}}^{\frac{π}{2}} cos^{2} (u) d u

\int_{1}^{4} (x^{2} + 1) d x

\int_{0}^{\frac{π}{4}} (sec (t))^{2} d t

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