Q. 7 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

$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}\sqrt{1+{\mathrm{x}}^2}$

Step-by-Step Solution

Answer

The Final Answer is $\frac{1}{2}\int \sqrt{\mathrm{t}}\mathrm{dt}=\frac{1}{3}{\left[1+{\mathrm{x}}^2\right]}^{\frac{3}{2}}+\mathrm{C}$

Step 1: Given information

The given function is $\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}\sqrt{1+{\mathrm{x}}^2}$.

Step 2: Calculations

The given function is $\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}\sqrt{1+{\mathrm{x}}^2}...\left(\mathrm{i}\right)$

${\mathrm{F}}^{\mathrm{\prime }}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{x}\right)$

Adding up the numbers (1)

$\int \text{x}\sqrt{1+{\text{x}}^2}\mathrm{dx}$

Multiplying and dividing by 2.

$\frac{1}{2}\int 2\mathrm{x}\sqrt{1+{\mathrm{x}}^2}\mathrm{dx}...\left(\mathrm{ii}\right)$

Assume that,$1+{\mathrm{x}}^2=\mathrm{t}$distinguishing in terms of x.

$2\mathrm{xdx}=\mathrm{dt}$

Replace the value in equation (2).

$\frac{1}{2}\int \sqrt{\mathrm{t}}\mathrm{dt}$

Now, integrate the obtained function.

$\begin{array}{r}\frac{1}{2}\int \sqrt{\mathrm{t}}\mathrm{dt}=\frac{1}{2}\frac{|\mathrm{t}{|}^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\mathrm{C}\\ \frac{1}{2}\int \sqrt{\mathrm{t}}\mathrm{dt}=\frac{1}{2}\times \left(\frac{2}{3}\right)[\mathrm{t}{]}^{\frac{3}{2}}+\mathrm{C}\\ \frac{1}{2}\int \sqrt{\mathrm{t}}\mathrm{dt}=\frac{1}{3}[\mathrm{t}{]}^{\frac{3}{2}}+\mathrm{C}\end{array}$

Where C denotes the integration constant.

Adding the value of t to the equation above.

$\frac{1}{2}\int \sqrt{\mathrm{t}}\mathrm{dt}=\frac{1}{3}{\left[1+{\mathrm{x}}^2\right]}^{\frac{3}{2}}+\mathrm{C}$