What is the integral from 1 to 3 of sqrt(5+4x-x^2) ?

The integral from 1 to 3 of sqrt(5+4x-x^2) is 9arcsin(1/3)+2sqrt(2)

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integral from 1 to 3 of sqrt(5+4x-x^2)

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

9 arcsin (\frac{1}{3}) + 22

Decimal

5.88695 \dots

Solution steps

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

Complete the square

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

= \int_{1}^{3} - (x - 2)^{2} + 9 d x

Apply u-substitution

= \int_{- 1}^{1} - u^{2} + 9 d u

Apply Trigonometric Substitution

= \int_{- a r c s i n (\frac{1}{3})}^{a r c s i n (\frac{1}{3})} 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_{- a r c s i n (\frac{1}{3})}^{a r c s i n (\frac{1}{3})} cos^{2} (v) d v

Rewrite using trig identities

= 9 \cdot \int_{- a r c s i n (\frac{1}{3})}^{a r c s i n (\frac{1}{3})} \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_{- a r c s i n (\frac{1}{3})}^{a r c s i n (\frac{1}{3})} 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_{- a r c s i n (\frac{1}{3})}^{a r c s i n (\frac{1}{3})} 1 d v + \int_{- a r c s i n (\frac{1}{3})}^{a r c s i n (\frac{1}{3})} cos (2 v) d v)

\int_{- a r c s i n (\frac{1}{3})}^{a r c s i n (\frac{1}{3})} 1 d v = 2 arcsin (\frac{1}{3})

\int_{- a r c s i n (\frac{1}{3})}^{a r c s i n (\frac{1}{3})} cos (2 v) d v = \frac{4 2}{9}

= 9 \cdot \frac{1}{2} (2 arcsin (\frac{1}{3}) + \frac{4 2}{9})

Simplify

9 \cdot \frac{1}{2} (2 arcsin (\frac{1}{3}) + \frac{4 2}{9}) : 9 arcsin (\frac{1}{3}) + 22

= 9 arcsin (\frac{1}{3}) + 22

What is the integral from 1 to 3 of sqrt(5+4x-x^2) ?
The integral from 1 to 3 of sqrt(5+4x-x^2) is 9arcsin(1/3)+2sqrt(2)