Find the area bounded by the given curves

Use definite integrals to find the area bounded by the given curves.

1. y= x^2 and y=2x

2. y^2 = 4x and x=1 (hint integrate with respect to y)

Step 1: Determine the intersection points and select the appropriate coordinates for the area of integral [e.g. if you are integrating the expression w/respect to x, then use x-coordinates]

Step 2: Graph both functions and determine the appropriate functions. If you are integrating the function w/respect to x, then you first determine the upper and lower functions. If integrating the function w/respect to y, then you determine the right and left functions.

Step 3: Set up the area integral and do the integration! Apply FTC [First Fundamental Theorem of Calculus] at the end.

E.g.

1. y = x² and y = 2x

Determine the x-coordinates of both functions.

x² = 2x
x² - 2x = 0
x(x - 2) = 0
x = {0,2}

Take a look at the graph in this link: http://www.wolframalpha.com/input/?i=y+%…

The upper function is y = 2x and the lower function is y = x². We set up the area integral as:

A = ∫(x = 0,2) upper - lower dx
= ∫(x = 0,2) 2x - x² dx

Then, evaluate the integral.

A = 2x²/2 - x³/3 | 0,2
= 2(2)²/2 - (2)³/3 - (2(0)²/2 - 0³/3) [FTC]
= 4 - 8/3
= 4/3

2. y² = 4x and x = 1

Since you want to integrate w/respect to y, determine the y-coordinates of the intersection points.

y² = 4(1)
y² = 4
y = ±2

Take a look at the graph here: http://www.wolframalpha.com/input/?i=y%C…

The left function is y² = 4x or x = y²/4 [Note that you need to solve for x] and the right function is x = 1. We can set up the integral as:

A = ∫(y = -2,2) 1 - y²/4 dy

Hence:

A = y - y³/12 | -2,2
= 2 - (2)³/12 - (-2 - (-2)³/12)
= 8/3

I hope this helps!