What is the integral from 1 to 5 of sqrt(-x^2+6x-5) ?

The integral from 1 to 5 of sqrt(-x^2+6x-5) is 2pi

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integral from 1 to 5 of sqrt(-x^2+6x-5)

\int_{1}^{5} - x^{2} + 6 x - 5 d x

2 π

Decimal Notation

6.28318 \dots

Solution steps

\int_{1}^{5} - x^{2} + 6 x - 5 d x

Complete the square

- x^{2} + 6 x - 5 : - (x - 3)^{2} + 4

= \int_{1}^{5} - (x - 3)^{2} + 4 d x

Apply u-substitution

: \int_{- 2}^{2} - u^{2} + 4 d u

= \int_{- 2}^{2} - u^{2} + 4 d u

Apply Trigonometric Substitution

: \int_{- \frac{π}{2}}^{\frac{π}{2}} 4 cos^{2} (v) d v

= \int_{- \frac{π}{2}}^{\frac{π}{2}} 4 cos^{2} (v) d v

Take the constant out:

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

= 4 \cdot \int_{- \frac{π}{2}}^{\frac{π}{2}} cos^{2} (v) d v

Rewrite using trig identities

= 4 \cdot \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{1 + cos ( 2 v )}{2} d v

Take the constant out:

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

= 4 \cdot \frac{1}{2} \cdot \int_{- \frac{π}{2}}^{\frac{π}{2}} 1 + cos (2 v) d v

Apply the Sum Rule:

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

= 4 \cdot \frac{1}{2} (\int_{- \frac{π}{2}}^{\frac{π}{2}} 1 d v + \int_{- \frac{π}{2}}^{\frac{π}{2}} cos (2 v) d v)

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

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

= 4 \cdot \frac{1}{2} (π + 0)

Simplify

4 \cdot \frac{1}{2} (π + 0) : 2 π

= 2 π

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What is the integral from 1 to 5 of sqrt(-x^2+6x-5) ?
The integral from 1 to 5 of sqrt(-x^2+6x-5) is 2pi