What is the integral of r^4(7-(r^5)/(10))^3 ?

The integral of r^4(7-(r^5)/(10))^3 is -1/20000 (70-r^5)^4+C

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integral of r^4(7-(r^5)/(10))^3

\int r^{4} (7 - \frac{r ^{5}}{10})^{3} d r

- \frac{1}{20000} (70 - r^{5})^{4} + C

Solution steps

\int r^{4} (7 - \frac{r ^{5}}{10})^{3} d r

Apply u-substitution

: \int - 2 u^{3} d u

= \int - 2 u^{3} d u

Take the constant out:

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

= - 2 \cdot \int u^{3} d u

Apply the Power Rule

: \frac{u ^{4}}{4}

= - 2 \cdot \frac{u ^{4}}{4}

Substitute back

u = 7 - \frac{r ^{5}}{10}

= - 2 \cdot \frac{( 7 - \frac{r ^{5}}{10} ) ^{4}}{4}

Simplify

- 2 \cdot \frac{( 7 - \frac{r ^{5}}{10} ) ^{4}}{4} : - \frac{1}{20000} (70 - r^{5})^{4}

= - \frac{1}{20000} (70 - r^{5})^{4}

Add a constant to the solution

= - \frac{1}{20000} (70 - r^{5})^{4} + C

y^{^{'}} = \frac{3 t ^{2} - e ^{t}}{2 y - 5}

\int 3 t (5 t^{2} - 3 t + 2) d t

n = 1 \sum \infty 3 n + 1

\int_{0}^{1} 1 + (2 x)^{2} d x

y = 3 + 5 x

What is the integral of r^4(7-(r^5)/(10))^3 ?
The integral of r^4(7-(r^5)/(10))^3 is -1/20000 (70-r^5)^4+C