Q. 8TF - 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.

$\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2\sqrt{1+{\mathrm{x}}^3}$

Step-by-Step Solution

Answer

The final integrate function is $\frac{1}{3}\int \sqrt{\mathrm{t}}\mathrm{dt}=\frac{2}{9}{\left[1+{\mathrm{x}}^3\right]}^{\frac{3}{2}}+\mathrm{C}$

Step 1: Given Information

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

Step 2. Integrate the given function.

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

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

Adding up the numbers (1).

$\int {\mathrm{x}}^2\sqrt{1+{\mathrm{x}}^3}\mathrm{dx}$

Multiplying and dividing by 3.

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

Assume that$1+{\mathrm{x}}^3=\mathrm{t}$ and distinguish in terms of x.

$3{\mathrm{x}}^3\mathrm{dx}=\mathrm{dt}$

Replace the value in equation (2).

$\frac{1}{3}\int \sqrt{\mathrm{t}}\mathrm{dt}$

Now it's time to integrate.

$\begin{array}{l}\frac{1}{3}\int \sqrt{\mathrm{t}}\mathrm{dt}=\frac{1}{3}\frac{{\left[\mathrm{t}\right]}^{\frac{1}{2}+1}}{{}^{\frac{1}{2}+1}}+\mathrm{C}\\ \frac{1}{3}\int \sqrt{\mathrm{t}}\mathrm{dt}=\frac{1}{3}\times \left(\frac{2}{3}\right){\left[\mathrm{t}\right]}^{\frac{3}{2}}+\mathrm{C}\\ \frac{1}{3}\int \sqrt{\mathrm{t}}\mathrm{dt}=\frac{2}{9}{\left[\mathrm{t}\right]}^{\frac{3}{2}}+\mathrm{C}\end{array}$

And, C denotes the integration constant.

Adding the value of t to the equation above.

$\frac{1}{3}\int \sqrt{\mathrm{t}}\mathrm{dt}=\frac{2}{9}{\left[1+{\mathrm{x}}^3\right]}^{\frac{3}{2}}+\mathrm{C}$