Q. 13 - Calculus

Question

Suppose $u\left(x\right)=x^2$. Calculate and compare the values of the following definite integrals:

${\int }_{-1}^5{u}^{2}du,{\int }_{x=-1}^{x=5}{u}^{2}du \mathrm{and} {\int }_{u(-1)}^{u\left(5\right)}{u}^{2}du$

Step-by-Step Solution

Answer

$$\begin{gathered} \mathrm{The} \mathrm{value} \mathrm{of} \mathrm{integral} {\int }_{-1}^5{u}^{2}du=\frac{1}{3}\left[{\left(5\right)}^3-{(-1)}^3\right]=42 \\ \mathrm{The} \mathrm{value} \mathrm{of} \mathrm{integral}{\int }_{x=-1}^{x=5}{u}^{2}du=\frac{1}{3}\left[{\left(5\right)}^6-{(-1)}^6\right]=5208 \\ \mathrm{The} \mathrm{value} \mathrm{of} \mathrm{integral}{\int }_{u(-1)}^{u\left(5\right)}{u}^{2}du=\frac{1}{3}\left[{\left(25\right)}^3-{\left(1\right)}^3\right]=5208 \end{gathered}$$

Step 1. Given Information

Suppose $u\left(x\right)=x^2$. Calculate and compare the values of the following definite integrals:

${\int }_{-1}^5{u}^{2}du,{\int }_{x=-1}^{x=5}{u}^{2}du and {\int }_{u(-1)}^{u\left(5\right)}{u}^{2}du$

Step 2. Firstly calculating the first definite integral.

$$\begin{gathered} {\int }_{-1}^5{u}^{2}du={\left[\frac{u^{2+1}}{2+1}\right]}_{-1}^5 \\ {\int }_{-1}^5{u}^{2}du={\left[\frac{u^3}{3}\right]}_{-1}^5 \\ {\int }_{-1}^5{u}^{2}du=\frac{1}{3}{\left[u^3\right]}_{-1}^5 \\ {\int }_{-1}^5{u}^{2}du=\frac{1}{3}\left[{\left(5\right)}^3-{(-1)}^3\right] \\ {\int }_{-1}^5{u}^{2}du=\frac{1}{3}\left[125-(-1)\right] \\ {\int }_{-1}^5{u}^{2}du=\frac{1}{3}\left[125+1\right] \\ {\int }_{-1}^5{u}^{2}du=\frac{126}{3} \\ {\int }_{-1}^5{u}^{2}du=42 \end{gathered}$$

Step 3. Now solving the second integral.

$$\begin{gathered} {\int }_{x=-1}^{x=5}{u}^{2}du={\left[\frac{u^{2+1}}{2+1}\right]}_{x=-1}^{x=5} \\ {\int }_{x=-1}^{x=5}{u}^{2}du={\left[\frac{u^3}{3}\right]}_{x=-1}^{x=5} \\ \mathrm{Since} u\left(x\right)=x^2 \\ {\int }_{x=-1}^{x=5}{u}^{2}du={\left[\frac{{\left(x^2\right)}^3}{3}\right]}_{x=-1}^{x=5} \\ {\int }_{x=-1}^{x=5}{u}^{2}du={\left[\frac{x^6}{3}\right]}_{x=-1}^{x=5} \\ {\int }_{x=-1}^{x=5}{u}^{2}du=\frac{1}{3}{\left[x^6\right]}_{x=-1}^{x=5} \\ {\int }_{x=-1}^{x=5}{u}^{2}du=\frac{1}{3}\left[{\left(5\right)}^6-{(-1)}^6\right] \\ {\int }_{x=-1}^{x=5}{u}^{2}du=\frac{1}{3}\left[15625-1\right] \\ {\int }_{x=-1}^{x=5}{u}^{2}du=\frac{1}{3}\left[15624\right] \\ {\int }_{x=-1}^{x=5}{u}^{2}du=\frac{15624}{3} \\ {\int }_{x=-1}^{x=5}{u}^{2}du=5208 \end{gathered}$$

Step 4. Now solving the integral third.

$$\begin{gathered} {\int }_{u(-1)}^{u\left(5\right)}{u}^{2}du={\int }_{{(-1)}^2}^{{\left(5\right)}^2}{u}^{2}du \\ {\int }_{u(-1)}^{u\left(5\right)}{u}^{2}du={\int }_1^{25}{u}^{2}du \\ {\int }_{u(-1)}^{u\left(5\right)}{u}^{2}du={\left[\frac{u^{2+1}}{2+1}\right]}_1^{25} \\ {\int }_{u(-1)}^{u\left(5\right)}{u}^{2}du={\left[\frac{u^3}{3}\right]}_1^{25} \\ {\int }_{u(-1)}^{u\left(5\right)}{u}^{2}du=\frac{1}{3}{\left[u^3\right]}_1^{25} \\ {\int }_{u(-1)}^{u\left(5\right)}{u}^{2}du=\frac{1}{3}\left[{\left(25\right)}^3-{\left(1\right)}^3\right] \\ {\int }_{u(-1)}^{u\left(5\right)}{u}^{2}du=\frac{1}{3}\left[15625-1\right] \\ {\int }_{u(-1)}^{u\left(5\right)}{u}^{2}du=\frac{1}{3}\left[15624\right] \\ {\int }_{u(-1)}^{u\left(5\right)}{u}^{2}du=\frac{15624}{3} \\ {\int }_{u(-1)}^{u\left(5\right)}{u}^{2}du=5208 \end{gathered}$$