Integration applied to Areas and Volumes

Definite integrals

Definite Integrals

 $\int_a^bf\left(x\right)\ dx=\left(g\left(x\right)\right)_a^b$  
 $=g\left(b\right)-g\left(a\right)$  
where:
*g(x) is the integrated function of f(x) 
* b and a are integers (negative or positive numbers) and b has a greater value than a.
*g(b) is when the number b is subbed into the integrated function 
*g(a) is when the number a is subbed into the integrated function
*+c is not shown because it cancels out 

A numerical answer is produced.

Examples

Find the value of 
1.  $\int_{-2}^4\left(x^3+5x\right)dx$  
 $=\left(\frac{x^4}{4}+\frac{5x^2}{2}\right)_{-2}^4$  
 $=\left(\frac{\left(4\right)^4}{4}+\frac{5\left(4\right)^2}{2}\right)-\left(\frac{\left(-2\right)^4}{4}+\frac{5\left(-2\right)^2}{2}\right)$  
 $=104-14$  
 $=90$  

2.  $\int_1^3\left(3x^2-1\right)dx$                   

 $=\left(\frac{3x^3}{3}-x\right)_1^3$  
 $=\left(x^3-x\right)_1^3$  
 $=\left(\left(3\right)^3-3\right)-\left(\left(1\right)^3-1\right)$  
 $=24-0$  
 $=24$  

Area under a curve

 $Area=\int_{x_2}^{x_1}y\ dx$  
where:
*  $x_1and\ x_2$  are the 2 x value boundaries under the curve 
*y is the equation of the curve 
*the value obtained is in square units to represent area.

There are 4 boundaries: 

1. curve
2.  $x_1$  
3.  $x_2$  
4. x axis 

Examples

1. The graph shows the equation  $y=x^2$  .   Find the area under the curve between x=2 and x=5 and the x axis.

 $A=\int_2^5x^2dx$  
 $=\left(\frac{x^3}{3}\right)_2^5$  
 $=\left(\frac{\left(5\right)^3}{3}\right)-\left(\frac{\left(2\right)^3}{3}\right)$  
 $=\frac{125}{3}-\frac{8}{3}$  
 $=\frac{117}{3}$  
 $=39\ squ\ units$  

 

2. Find the area of the region enclosed by the curve  $y=3x^2+8x+6,\$  the x axis and the lines x=-2 and x=1.

 $A=\int_{x_2}^{x_1}y\ dx$  
 $=\int_{-2}^1\left(3x^2+8x+6\right)dx$  
 $=\left(x^3+4x^2+6x\right)_{-2}^1$  
 $=\left(\left(1\right)^3+4\left(1\right)^2+6\left(-2\right)\right)-\left(\left(-2\right)^3+4\left(-2\right)^2+6\left(-2\right)\right)$  
 $=11-\left(-4\right)$  
 $15\ squ.\ units$  

3. Find the area of the region enclosed by the curve  $y=2x+3x^{\frac{1}{2}}$   the x axis and the lines x=1 and x=4.

 $A=\int_{x_2}^{x_1}y\ dx$  
 $=\int_1^4\left(2x+3x^{\frac{1}{2}}\right)$  
 $=\left(x^2+2x^{\frac{3}{2}}\right)_1^4\ \ \ \ \ \ \ \ \ \ \left(3x^{\frac{3}{2}}\times\frac{2}{3}\right)$  
 $=\left(\left(4\right)^2+2\left(4\right)^{\frac{3}{2}}\right)-\left(\left(1\right)^2+2\left(1\right)^{\frac{3}{2}}\right)$  
 $=32-3$  
 $=29\ squ\ units$  

4. Find the area bounded by the curve  $y=5\sin x+\cos x$  , the lines x=0 and  $x=\frac{\pi}{2}$  and the x axis.

 $A=\int_{x_2}^{x_1}y\ dx$  
 $=\int_0^{\frac{\pi}{2}}\left(5\sin x+\cos x\right)dx$  
 $=\left(-5\cos x+\sin x\right)_0^{\frac{\pi}{2}}$  
 $=\left(-5\cos\left(\frac{\pi}{2}\right)+\sin\left(\frac{\pi}{2}\right)\right)-\left(-5\cos\left(0\right)+\sin\left(0\right)\right)$  
 $1-\left(-5\right)$  
 $6\ squ\ units$  

Volume of Solid formed when a Plane Area  is rotated about the x axis.

Formula:

 $V=\pi\int_{x_2}^{x_1}y^2dx$  
where:
*PI is a "constant" that remains outside the integral sign & brackets until final answer
*  $x_1and\ x_2$  are the 2 x values of the boundary/ the upper and lower limits
*  $y^2$   is the equation of the curve squared
 *the value obtained is in cubic units to represent volume

Examples

1. Find the volume of the solid generated when the region bounded by the line  $y=x+2$  , the axis and the lines x=2 and x=5 is rotated  $360\degree$  about the x axis.

 $V=\pi\int_{x_2}^{x_1}y^2dx$                               $y=\left(x+2\right)\therefore y^2\left(x+2\right)^2=x^2+4x+4$  
 $=\pi\int_2^5\left(x^2+4x+4\right)dx$  
 $=\pi\left(\frac{x^3}{3}+2x^2+4x\right)_2^5$  
 $=\pi\left(\left(\frac{\left(5\right)^3}{3}+2\left(5\right)^2+4\left(5\right)\right)-\left(\frac{\left(2\right)^3}{3}+2\left(2\right)^2+4\left(2\right)\right)\right)$  
 $=\pi\left(\frac{335}{3}-\frac{56}{3}\right)$  
 $=\pi\left(\frac{279}{3}_{ }\right)$  
 $=93\pi\ cubic\ units$  

2.Find the volume generated when the curve  $x^2+y^2=25$  is rotated 360 degrees between x=1 and x=4.

 $y^2=25-x^2$  
 $V=\pi\int_{x_2}^{x_1}y^2dx$  
 $=\pi\int_1^4\left(25-x^2\right)dx$  
 $=\pi\left(25x-\frac{x^3}{3}\right)_1^4$  
 $=\pi\left(\left(25\left(4\right)-\frac{\left(4\right)^3}{3}\right)-\left(25\left(1\right)-\frac{\left(1\right)^3}{3}\right)\right)$  
 $=\pi\left(\frac{236}{3}-\frac{74}{3}\right)$  
 $=\pi\left(\frac{162}{3}\right)$  
 $=54\pi\ cubic\ units$  

3. The region R is bounded by the part of the curve  $y=\left(x-2\right)^{\frac{3}{2}}$  for which  $2\le x\le4$  , the x axis and the line x=4. Find in terms of   $\pi$  , the volume of the solid obtained when R is rotated 4 right angles about the x axis.

 $y^2=\left(\left(x-2\right)^{\frac{3}{2}}\right)^2=\left(x-2\right)^3$  
 $V=\pi\int_{x_2}^{x_1}y^2dx$  
 $=\pi\int_2^4\left(\left(x-2\right)^3\right)dx$  
 $=\pi\left(\frac{\left(x-2\right)^4}{4}\right)_2^4$  
 $=\pi\left(\left(\frac{\left(4-2\right)^4}{4}\right)-\left(\frac{\left(2-2\right)^4}{4}\right)\right)$  
 $=\pi\left(4-0\right)$  
 $=4\pi$  cubic units

4.Find the volume generated when the curve  $y=\sqrt{\cos\ x}$  is rotated 360 degrees between x=0 and x= $\frac{\pi}{2}$  .

 $y^2=\left(\left(\cos x\right)^{\frac{1}{2}}\right)^2=\cos\ x$  
 $V=\pi\int_{x_2}^{x_1}y^2dx$  
 $=\pi\int_0^{\frac{\pi}{2}}\left(\cos x\right)_{ }^{ }$  
 $=\pi\left(\sin x\right)_0^{\frac{\pi}{2}}$  
 $=\pi\left(\left(\sin\left(\frac{\pi}{2}\right)\right)-\left(\sin\left(0\right)\right)\right)$  
 $=\pi\left(1-0\right)$  
 $=\pi$  cubic units