We have an answer to "Find the indefinite integral. ∫x8−xdx"

High School
Calculus and Analysis
Calculus 2
Applications of integrals

Mary Keefe

Answered question

2021-12-28

Find the indefinite integral.

\int x \sqrt{8 - x} d x

Answer & Explanation

Bob Huerta

Beginner2021-12-29Added 41 answers

Step 1
To solve this indefinite integral, we have to use the substitution method.
Substitute 8-x = u, that implies -dx = du, and x = 8 - u.
By substituting these values in the above integral,

= - \int (8 - u) \sqrt{u} d u

= - \int 8 u^{\frac{1}{2}} - u^{\frac{3}{2}} d u

= \frac{2}{5} u^{\frac{5}{2}} - \frac{16}{3} u^{\frac{3}{2}} + C

Step 2
By resubstituting value u = (8 - x)

= \frac{2}{5} {(8 - x)}^{\frac{5}{2}} - \frac{16}{3} {(8 - x)}^{\frac{3}{2}} + C

This is the required integral.

Joseph Fair

Beginner2021-12-30Added 34 answers

\int x \sqrt{8 - x} d x

= \int (u^{\frac{3}{2}} - 8 \sqrt{u}) d u

= \int u^{\frac{3}{2}} d u - 8 \int \sqrt{u} d u

\int u^{\frac{3}{2}} d u

= \frac{2 u^{\frac{5}{2}}}{5}

\int \sqrt{u} d u

= \frac{2 u^{\frac{3}{2}}}{3}

\int u^{\frac{3}{2}} d u - 8 \int \sqrt{u} d u

= \frac{2 u^{\frac{5}{2}}}{5} - \frac{16 u^{\frac{3}{2}}}{3}

= \frac{2 {(8 - x)}^{\frac{5}{2}}}{5} - \frac{16 {(8 - x)}^{\frac{3}{2}}}{3}

\int x \sqrt{8 - x} d x

= \frac{2 {(8 - x)}^{\frac{5}{2}}}{5} - \frac{16 {(8 - x)}^{\frac{3}{2}}}{3} + C

= \frac{2 (- 3 x - 16) {(8 - x)}^{\frac{3}{2}}}{15} + C

karton

Expert2022-01-04Added 613 answers

$\begin{matrix} x \sqrt{8 - x} d x \\ \int t \sqrt{t} - 8 \sqrt{t} d t \\ \int t \times t^{\frac{1}{2}} - 8 t^{\frac{1}{2}} d t \\ \int t^{\frac{3}{2}} - 8 t^{\frac{1}{2}} d t \\ \int t^{\frac{3}{2}} d t - \int 8 t^{\frac{1}{2}} d t \\ \frac{2 t^{2} \sqrt{t}}{5} - \frac{16 t \sqrt{t}}{3} \\ \frac{2 (8 - x)^{2} \times \sqrt{8 - x}}{5} - \frac{16 (8 - x) \sqrt{8 - x}}{3} \\ \frac{2 \sqrt{8 - x} (64 - 16 x + x^{2})}{5} - \frac{16 (8 - x) \sqrt{8 - x}}{3} \end{matrix}$
Answer:
$\frac{2 \sqrt{8 - x} (64 - 16 x + x^{2})}{5} - \frac{16 (8 - x) \sqrt{8 - x}}{3} + C$

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