What is the integral of (x/2)^4 ?

The integral of (x/2)^4 is (x^5)/(16)-(x^5)/(20)+C

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integral of (x/2)^4

\int (\frac{x}{2})^{4} d x

\frac{x ^{5}}{16} - \frac{x ^{5}}{20} + C

Solution steps

\int (\frac{x}{2})^{4} d x

Apply Integration By Parts

= \frac{x ^{5}}{16} - \int \frac{x ^{4}}{4} d x

\int \frac{x ^{4}}{4} d x = \frac{x ^{5}}{20}

= \frac{x ^{5}}{16} - \frac{x ^{5}}{20}

Add a constant to the solution

= \frac{x ^{5}}{16} - \frac{x ^{5}}{20} + C

y^{^{'}} = \frac{y}{x}

\frac{\partial}{\partial y} (\frac{1}{ln ( 10 ) ( x - y )})

\frac{d y}{d t} + 3 y + e^{t} + 3 = 0

\int (6 sec^{2} (x)) d x

x^{3} + 3 y - x \frac{d y}{d x} = 0

What is the integral of (x/2)^4 ?
The integral of (x/2)^4 is (x^5)/(16)-(x^5)/(20)+C