What is the integral from 1 to 5 of sqrt(9-6x+x^2) ?

The integral from 1 to 5 of sqrt(9-6x+x^2) is 4

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integral from 1 to 5 of sqrt(9-6x+x^2)

\int_{1}^{5} 9 - 6 x + x^{2} d x

4

Solution steps

\int_{1}^{5} 9 - 6 x + x^{2} d x

Factor

9 - 6 x + x^{2} : (x - 3)^{2}

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

Apply exponent rule:

a^{2} = ∣ a ∣

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

Eliminate Absolutes

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

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

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

= 2 + 2

Simplify

= 4

\int_{4}^{9} \frac{9 x + 1}{x} d x

\int_{2}^{5} [x^{2} - 2 x + 3] d x

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

\int_{0}^{3} (5 x + 1) d x

\int_{0}^{\frac{π}{2}} cos^{13} (x) d x

What is the integral from 1 to 5 of sqrt(9-6x+x^2) ?
The integral from 1 to 5 of sqrt(9-6x+x^2) is 4