Post #32

Finding the Area and volume under a curve

Int (Top Equation)-(Bottom Equation) on the interval [a,b]

You can find a and b by setting the equations equal to each other and solving because a and b are their intersections.

Two rules needed to know: If the area is on the y then a and b need to be y values and solved for x. If the area is on the x then a and b need to be x values and solved for y.

Example:

Find the area of the rigion bounded by f(x)=2-x^2 and g(x)=x.

2-x^2=x 2-x^2-x=0 (-x-1)(x+2) x=-1 x=2

int (Top)-(Bottom) dx [-1,2]

To find the top and bottom equation just graph them on your graphing calculator. You'll see that 2-x^2 is on top with x on the bottom.

int (2-x^2)-(x) dx [-1,2]

int (2-x^2-x) = 2x-(1/3)x^3-(1/2)x^2

Solve like an ordinary definite integral.

2(2)-(1/3)((2)^3)-(1/2)((2)^2)-[2(-1)-(1/3)((-1)^3)-(1/2)((-1)^2)]=(3/2)

Volume is little different because there are two ways to find the volume of a region, depending on the region itself. The two methods are discs and washers.

Discs: (π)int [R(x)]² dx [a,b]

Example:

(π)int √(sinx)² dx [0,π]

(π)int sinx dx [0,π]

(π)(-cosx) [0,π] -cos(π)-(-cos(0))

π(1+1)=2π

Washers: (π)int (Top equation)²-(Bottom equation)² dx [a,b]

Example:

√(x) and (x²)

(π)int (√(x))² - ((x²))² [0,1]

((1/2)x^2) - ((1/5)x^5) 1/2(1)-(1/5)(1) - [1/2(0) -(1/5)(0)]= 3/10

(π)(3/10)= (3π)/10