Q. 9 TF - Calculus

Question

Finding antiderivatives by undoing the chain rule: For each function f that follows, find a function F with the property that $\mathrm{F}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{x}\right)$. You may have to guess and check to find such a function.

$\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{{\left(2-5\mathrm{x}\right)}^3}$

Step-by-Step Solution

Answer

The final integrate function is $-\frac{1}{5}\int \frac{1}{{\left(\mathrm{t}\right)}^3}\mathrm{dt}=\frac{1}{10{\left(2-5\mathrm{x}\right)}^2}+\mathrm{C}$

Step 1. Given Information.

The given function is $\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{{\left(2-5\mathrm{x}\right)}^3}$.

Step 2. Find the integration of the given function.

The given function is $\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{{\left(2-5\mathrm{x}\right)}^3}....\left(\mathrm{i}\right)$

$\mathrm{F}\text{'}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{x}\right)$

Adding up the numbers (1).

$\int \frac{1}{(2-5x{)}^3}dx$

Multiplying and dividing by 5.

$\frac{1}{5}\int \frac{5}{(2-5\mathrm{x}{)}^3}\mathrm{dx}...\left(\mathrm{ii}\right)$

Assume that$2-5\mathrm{x}=\mathrm{t}$ distinguishing in terms of x.

$5\mathrm{xdx}=-\mathrm{dt}$

Replace the value in equation (2).

$-\frac{1}{5}\int \frac{1}{{\displaystyle {\left(\mathrm{t}\right)}^3}}\mathrm{dt}$

Now integrate the obtained function.

$\begin{array}{l}-\frac{1}{5}\int \frac{1}{{\left(\mathrm{t}\right)}^3}\mathrm{dt}=-\frac{1}{5}\int {\left(\mathrm{t}\right)}^{-3}\mathrm{dt}\\ -\frac{1}{5}\int \frac{1}{{\left(\mathrm{t}\right)}^3}\mathrm{dt}=-\frac{1}{5}\frac{{\left[\text{t}\right]}^{-3+1}}{-3+1}+\text{C}\\ -\frac{1}{5}\int \frac{1}{{\left(\mathrm{t}\right)}^3}\mathrm{dt}=\frac{1}{5}\times \left(\frac{1}{2}\right){\left[\mathrm{t}\right]}^{-2}+\text{C}\\ -\frac{1}{5}\int \frac{1}{{\left(\mathrm{t}\right)}^3}\mathrm{dt}=\frac{1}{10{\left(\mathrm{t}\right)}^2}+\text{C}\end{array}$

C denotes the integration constant.

Adding the value of t to the equation above.

$-\frac{1}{5}\int \frac{1}{{\left(\mathrm{t}\right)}^3}\mathrm{dt}=\frac{1}{10{\left(2-5\mathrm{x}\right)}^2}+\mathrm{C}$