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

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

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

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

\frac{π}{2}

Decimal Notation

1.57079 \dots

Solution steps

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

Apply Trigonometric Substitution

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

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

Rewrite using trig identities

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

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

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

= \frac{1}{2} (π + 0)

Simplify

\frac{1}{2} (π + 0) : \frac{π}{2}

= \frac{π}{2}

f (x) = x^{2} x - 5 x

\frac{d}{d x} (3 e^{\frac{x}{4}})

x^{2} - 2 x + 3

\frac{\partial}{\partial u} (v)

y = 7 x - x^{2}

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