What is the integral from-3 to 3 of-sqrt(9-x^2) ?

The integral from-3 to 3 of-sqrt(9-x^2) is -(9pi)/2

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integral from-3 to 3 of-sqrt(9-x^2)

\int_{- 3}^{3} - 9 - x^{2} d x

- \frac{9 π}{2}

Decimal Notation

- 14.13716 \dots

Solution steps

\int_{- 3}^{3} - 9 - x^{2} d x

Take the constant out:

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

= - \int_{- 3}^{3} 9 - x^{2} d x

Apply Trigonometric Substitution

: \int_{- \frac{π}{2}}^{\frac{π}{2}} 9 cos^{2} (u) d u

= - \int_{- \frac{π}{2}}^{\frac{π}{2}} 9 cos^{2} (u) d u

Take the constant out:

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

= - 9 \cdot \int_{- \frac{π}{2}}^{\frac{π}{2}} cos^{2} (u) d u

Rewrite using trig identities

= - 9 \cdot \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

= - 9 \cdot \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

= - 9 \cdot \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

= - 9 \cdot \frac{1}{2} (π + 0)

Simplify

- 9 \cdot \frac{1}{2} (π + 0) : - \frac{9 π}{2}

= - \frac{9 π}{2}

\int (x - 3)^{- 1} d x

\int \frac{1 + x ^{8}}{x ^{13}} d x

1 - 6 x, at x = - 1

- \frac{2}{( s + 2 ) ^{2}}

\int sin (3)

What is the integral from-3 to 3 of-sqrt(9-x^2) ?
The integral from-3 to 3 of-sqrt(9-x^2) is -(9pi)/2