What is the integral from-1 to 5 of sqrt(5+4x-x^2) ?

The integral from-1 to 5 of sqrt(5+4x-x^2) is (9pi)/2

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integral from-1 to 5 of sqrt(5+4x-x^2)

\int_{- 1}^{5} 5 + 4 x - x^{2} d x

\frac{9 π}{2}

Decimal

14.13716 \dots

Solution steps

\int_{- 1}^{5} 5 + 4 x - x^{2} d x

Complete the square

5 + 4 x - x^{2} : - (x - 2)^{2} + 9

= \int_{- 1}^{5} - (x - 2)^{2} + 9 d x

Apply u-substitution

: \int_{- 3}^{3} - u^{2} + 9 d u

= \int_{- 3}^{3} - u^{2} + 9 d u

Apply Trigonometric Substitution

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

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

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} (v) d v

Rewrite using trig identities

= 9 \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

= 9 \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

= 9 \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

= 9 \cdot \frac{1}{2} (π + 0)

Simplify

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

= \frac{9 π}{2}

\int_{π}^{\frac{3 π}{2}} sin (x) d x

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

What is the integral from-1 to 5 of sqrt(5+4x-x^2) ?
The integral from-1 to 5 of sqrt(5+4x-x^2) is (9pi)/2