Q.1 TB - Calculus

Question

Calculate each of the following definite integrals, using integration techniques and the Fundamental Theorem of Calculus.

$$\begin{gathered} {\int }_0^{10}4\mathrm{\pi }\left(11.2+0.072{\mathrm{x}}^2\right)dx \\ {\int }_0^4\left(62.4\mathrm{\pi }\right){\left(3.5\right)}^2\left(4-\mathrm{y}\right)dy \\ {\int }_0^{4}62.4\left(10+\frac{5}{2}y\right)\left(8-y\right)dy \\ \mathrm{\pi }{\int }_0^{100}\left(0.75x+250\right)\left(200-\mathrm{x}\right)dx \\ \mathrm{\pi }{\int }_0^6\left(-2\mathrm{y}+1.12\right)\left(100-\frac{25}{2}y\right)dy \end{gathered}$$

Step-by-Step Solution

Answer

The integrals of the expressions are:

$$\begin{gathered} {\int }_0^{10}4\mathrm{\pi }\left(11.2+0.072{\mathrm{x}}^2\right)dx=544\mathrm{\pi } \\ {\int }_0^4\left(62.4\mathrm{\pi }\right){\left(3.5\right)}^2\left(4-\mathrm{y}\right)dy=6115.2\mathrm{\pi } \\ {\int }_0^{4}62.4\left(10+\frac{5}{2}y\right)\left(8-y\right)dy=21632 \\ \mathrm{\pi }{\int }_0^{100}\left(0.75\mathrm{x}+250\right)\left(200-\mathrm{x}\right)d\mathrm{x}=4250000\mathrm{\pi } \\ \mathrm{\pi }{\int }_0^6\left(-2\mathrm{y}+1.12\right)\left(100-\frac{25}{2}\mathrm{y}\right)d\mathrm{y}=-1380\mathrm{\pi } \end{gathered}$$

Step 1. Given Information

Given to calculate each of the following integral:

$$\begin{gathered} a.{\int }_0^{10}4\mathrm{\pi }\left(11.2+0.072{\mathrm{x}}^2\right)dx \\ b.{\int }_0^4\left(62.4\mathrm{\pi }\right){\left(3.5\right)}^2\left(4-\mathrm{y}\right)dy \\ c.{\int }_0^{4}62.4\left(10+\frac{5}{2}y\right)\left(8-y\right)dy \\ d.\mathrm{\pi }{\int }_0^{100}\left(0.75x+250\right)\left(200-\mathrm{x}\right)dx \\ e.\mathrm{\pi }{\int }_0^6\left(-2\mathrm{y}+1.12\right)\left(100-\frac{25}{2}y\right)dy \end{gathered}$$

Part (a) Step 1. Calculating the integral

Solving the integral:

$$\begin{gathered} {\int }_0^{10}4\mathrm{\pi }\left(11.2+0.072{\mathrm{x}}^2\right)dx=4\mathrm{\pi }{\int }_0^{10}\left(11.2+0.072{\mathrm{x}}^2\right)dx \\ {\int }_0^{10}4\mathrm{\pi }\left(11.2+0.072{\mathrm{x}}^2\right)dx=\left(4\mathrm{\pi }\right){\left(11.2x+\frac{0.072{\mathrm{x}}^3}{3}\right)}_0^{10} \\ {\int }_0^{10}4\mathrm{\pi }\left(11.2+0.072{\mathrm{x}}^2\right)dx=\left(4\mathrm{\pi }\right)\left(\left(11.2\left(10\right)+\frac{0.072{\left(10\right)}^3}{3}\right)-\left(11.2\left(0\right)+\frac{0.072{\left(0\right)}^3}{3}\right)\right) \\ {\int }_0^{10}4\mathrm{\pi }\left(11.2+0.072{\mathrm{x}}^2\right)dx=\left(4\mathrm{\pi }\right)\left(\left(112+24\right)-\left(0+0\right)\right) \\ {\int }_0^{10}4\mathrm{\pi }\left(11.2+0.072{\mathrm{x}}^2\right)dx=\left(4\mathrm{\pi }\right)\left(136\right) \\ {\int }_0^{10}4\mathrm{\pi }\left(11.2+0.072{\mathrm{x}}^2\right)dx=544\mathrm{\pi } \end{gathered}$$

Part (b) Step 1. Calculating the integral

Solving the integral:

$$\begin{gathered} {\int }_0^4\left(62.4\mathrm{\pi }\right){\left(3.5\right)}^2\left(4-\mathrm{y}\right)dy=\left(62.4\mathrm{\pi }\right){\left(3.5\right)}^2{\int }_0^4\left(4-\mathrm{y}\right)dy \\ {\int }_0^4\left(62.4\mathrm{\pi }\right){\left(3.5\right)}^2\left(4-\mathrm{y}\right)dy=\left(62.4\mathrm{\pi }\right){\left(3.5\right)}^2{\left(4x-\frac{{\mathrm{y}}^2}{2}\right)}_0^4 \\ {\int }_0^4\left(62.4\mathrm{\pi }\right){\left(3.5\right)}^2\left(4-\mathrm{y}\right)dy=\left(62.4\mathrm{\pi }\right){\left(3.5\right)}^2\left(\left(4\left(4\right)-\frac{{\left(4\right)}^2}{2}\right)-\left(4\left(0\right)-\frac{{\left(0\right)}^2}{2}\right)\right) \\ {\int }_0^4\left(62.4\mathrm{\pi }\right){\left(3.5\right)}^2\left(4-\mathrm{y}\right)dy=\left(62.4\mathrm{\pi }\right)\left(12.25\right)\left(\left(16-8\right)-\left(0-0\right)\right) \\ {\int }_0^4\left(62.4\mathrm{\pi }\right){\left(3.5\right)}^2\left(4-\mathrm{y}\right)dy=\left(62.4\mathrm{\pi }\right)\left(12.25\right)\left(8\right) \\ {\int }_0^4\left(62.4\mathrm{\pi }\right){\left(3.5\right)}^2\left(4-\mathrm{y}\right)dy=6115.2\mathrm{\pi } \end{gathered}$$

Part (c) Step 1. Calculating the integral

Solving the integral:

$$\begin{gathered} {\int }_0^{4}62.4\left(10+\frac{5}{2}y\right)\left(8-y\right)dy=62.4{\int }_0^4\left(80+10y-\frac{5}{2}{y}^2\right)dy \\ {\int }_0^{4}62.4\left(10+\frac{5}{2}y\right)\left(8-y\right)dy=62.4{\left(80y+\frac{{10y}^2}{2}-\frac{5y^3}{6}\right)}_0^4 \\ {\int }_0^{4}62.4\left(10+\frac{5}{2}y\right)\left(8-y\right)dy=62.4\left(\left(80\left(4\right)+\frac{{10\left(4\right)}^2}{2}-\frac{5{\left(4\right)}^3}{6}\right)-\left(80\left(0\right)+\frac{{10\left(0\right)}^2}{2}-\frac{5{\left(0\right)}^3}{6}\right)\right) \\ {\int }_0^{4}62.4\left(10+\frac{5}{2}y\right)\left(8-y\right)dy=62.4\left(\left(320+80-\frac{160}{3}\right)-\left(0+0-0\right)\right) \\ {\int }_0^{4}62.4\left(10+\frac{5}{2}y\right)\left(8-y\right)dy=62.4\left(\frac{1040}{3}\right) \\ {\int }_0^{4}62.4\left(10+\frac{5}{2}y\right)\left(8-y\right)dy=21632 \end{gathered}$$

Part (d) Step 1. Calculating the integral

Solving the integral:

$$\begin{gathered} \mathrm{\pi }{\int }_0^{100}\left(0.75x+250\right)\left(200-\mathrm{x}\right)dx=\mathrm{\pi }{\int }_0^{100}\left(50000-100\mathrm{x}-0.75{\mathrm{x}}^2\right)dx \\ \mathrm{\pi }{\int }_0^{100}\left(0.75x+250\right)\left(200-\mathrm{x}\right)dx=\mathrm{\pi }{\left(50000x-50{\mathrm{x}}^2-0.25{\mathrm{x}}^3\right)}_0^{100} \\ \mathrm{\pi }{\int }_0^{100}\left(0.75\mathrm{x}+250\right)\left(200-\mathrm{x}\right)d\mathrm{x}=\mathrm{\pi }\left(\left(50000\left(100\right)-50{\left(100\right)}^2-0.25{\left(100\right)}^3\right)-\left(50000\left(0\right)-50{\left(0\right)}^2-0.25{\left(0\right)}^3\right)\right) \\ \mathrm{\pi }{\int }_0^{100}\left(0.75\mathrm{x}+250\right)\left(200-\mathrm{x}\right)d\mathrm{x}=\mathrm{\pi }\left(\left(5000000-500000-250000\right)-\left(0-0-0\right)\right) \\ \mathrm{\pi }{\int }_0^{100}\left(0.75\mathrm{x}+250\right)\left(200-\mathrm{x}\right)d\mathrm{x}=4250000\mathrm{\pi } \end{gathered}$$

Part (e) Step 1. Calculating the integral

Solving the integral:

$$\begin{gathered} \mathrm{\pi }{\int }_0^6\left(-2\mathrm{y}+1.12\right)\left(100-\frac{25}{2}y\right)dy=\mathrm{\pi }{\int }_0^6\left(25{\mathrm{y}}^2-214\mathrm{y}+112\right)dy \\ \mathrm{\pi }{\int }_0^6\left(-2\mathrm{y}+1.12\right)\left(100-\frac{25}{2}y\right)dy=\mathrm{\pi }{\left(\frac{25{\mathrm{y}}^3}{3}-107{\mathrm{y}}^2+112x\right)}_0^6 \\ \mathrm{\pi }{\int }_0^6\left(-2\mathrm{y}+1.12\right)\left(100-\frac{25}{2}\mathrm{y}\right)d\mathrm{y}=\mathrm{\pi }\left(\left(\frac{25{\left(6\right)}^3}{3}-107{\left(6\right)}^2+112\left(6\right)\right)-\left(\frac{25{\left(0\right)}^3}{3}-107{\left(0\right)}^2+112\left(0\right)\right)\right) \\ \mathrm{\pi }{\int }_0^6\left(-2\mathrm{y}+1.12\right)\left(100-\frac{25}{2}\mathrm{y}\right)d\mathrm{y}=\mathrm{\pi }\left(\left(1800-3852+672\right)-\left(0-0+0\right)\right) \\ \mathrm{\pi }{\int }_0^6\left(-2\mathrm{y}+1.12\right)\left(100-\frac{25}{2}\mathrm{y}\right)d\mathrm{y}=-1380\mathrm{\pi } \end{gathered}$$

Step 2. Conclusion

The integrals of the given expressions are:

$$\begin{gathered} {\int }_0^{10}4\mathrm{\pi }\left(11.2+0.072{\mathrm{x}}^2\right)dx=544\mathrm{\pi } \\ {\int }_0^4\left(62.4\mathrm{\pi }\right){\left(3.5\right)}^2\left(4-\mathrm{y}\right)dy=6115.2\mathrm{\pi } \\ {\int }_0^{4}62.4\left(10+\frac{5}{2}y\right)\left(8-y\right)dy=21632 \\ \mathrm{\pi }{\int }_0^{100}\left(0.75\mathrm{x}+250\right)\left(200-\mathrm{x}\right)d\mathrm{x}=4250000\mathrm{\pi } \\ \mathrm{\pi }{\int }_0^6\left(-2\mathrm{y}+1.12\right)\left(100-\frac{25}{2}\mathrm{y}\right)d\mathrm{y}=-1380\mathrm{\pi } \end{gathered}$$