What is the integral from-5 to 5 of sqrt(5^2-x^2) ?

The integral from-5 to 5 of sqrt(5^2-x^2) is (25pi)/2

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integral from-5 to 5 of sqrt(5^2-x^2)

\int_{- 5}^{5} 5^{2} - x^{2} d x

\frac{25 π}{2}

Decimal Notation

39.26990 \dots

Solution steps

\int_{- 5}^{5} 5^{2} - x^{2} d x

Apply Trigonometric Substitution

: \int_{- \frac{π}{2}}^{\frac{π}{2}} 25 cos^{2} (u) d u

= \int_{- \frac{π}{2}}^{\frac{π}{2}} 25 cos^{2} (u) d u

Take the constant out:

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

= 25 \cdot \int_{- \frac{π}{2}}^{\frac{π}{2}} cos^{2} (u) d u

Rewrite using trig identities

= 25 \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

= 25 \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

= 25 \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

= 25 \cdot \frac{1}{2} (π + 0)

Simplify

25 \cdot \frac{1}{2} (π + 0) : \frac{25 π}{2}

= \frac{25 π}{2}

\int_{2}^{3} (x - 3) e^{2 x} d x

\int_{1}^{4} \frac{1}{2 x} d x

\int_{- 1}^{1} (∣ x ∣ - 1) d x

\int_{3}^{4} x^{2} - 2 x - 3 d x

\int_{0}^{\frac{π}{2}} \frac{1}{1 + cos ( x )} d x

What is the integral from-5 to 5 of sqrt(5^2-x^2) ?
The integral from-5 to 5 of sqrt(5^2-x^2) is (25pi)/2