Recall that the area under a curve and above thex-axis can be computed by the definite integral. If we have two curves

( y = f(x) ) and ( y=g(x) )

such that

[ f(x) > g(x) nonumber]

then the area between them bounded by the horizontal lines (x = a) and (x = b) is

[ text{Area}=int_{c}^{b} left [ f(x) - g(x) right ] ;dx. nonumber]

Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor. Think about it: the area between the two curves is equal to the area under the top function minus the area that is under the bottom function. The people who don't like words can start paying attention again. We have two brand-spanking new formulas. Aren't you just itching to try them? Calculus Name Date Period ©R D220 U1x3Q CKsu XtSah JSWoLfYtGwVaRrUe8 LMLRCQ.e n 6Atl 8lR or Si6gSh 8tDsm crQehsVeBrLv Pe9d H.d Area Between Curves Practice For each problem, find the area of the region enclosed by the curves. 1) y = − x y = 2 x x = 0 x = 4 x y −8 −6 −4 −2 2 4 6 8 −8 −6 −4.

To remember this formula we write

Find the area between the curves ( y =0 ) and (y = 3 left( x^3-x right) ). When we graph the region, we see that the curves cross each other so that the top and bottom switch. In calculus, the evaluate the area between two curves, it is necessary to determine the difference of definite integrals of a function. The area between the two curves or function is defined as the definite integra l of one function (say f (x)) minus the definite integral of other functions (say g (x)).

[ text{Area}=int_{a}^{b}text{(Top-Bottom)};dx nonumber]

Example 1

Find the area between the curves ( y=x^2) and (y=x^3).

Solution

First we note that the curves intersect at the points ((0,0)) and ((1,1)). Then we see that

[ x^3 < x^2 nonumber]

in this interval. Hence the area is given by

[begin{align*} int_{0}^{1} left( x^2 - x^3 right) dx &= {left[ frac{1}{3}x^3 - frac{1}{4}x^4 right]}_0^1 &= dfrac{1}{3} - dfrac{1}{4} &= dfrac{1}{12}. end{align*}]

Area Bounded by Two Functions of (y)

Example 2

Find the area between the curves ( x = 1 - y^2 ) and ( x = y^2-1 ).

Solution

Here the curves bound the region from the left and the right.

We use the formula

[ text{Area}=int_{c}^{b}text{(Right-Left)};dy. nonumber]

For our example:

7.5 Area Between Curves Ap Calculus Formulas

[begin{align*} int_{-1}^{1}big[ (1-y^2)-(y^2-1) big] dy &= int_{-1}^{1}(2-y^2) dy &= left(2y-dfrac{2}{3}y^3right]_{-1}^1 &=big(2-dfrac{2}{3}big)-big(-2-dfrac{2}{3} big) &= dfrac{8}{3}. end{align*}]

Example 3

Find the area between the curves ( y =0 ) and (y = 3 left( x^3-x right) ).

Solution

7.5 Area Between Curves Ap Calculus Calculator

When we graph the region, we see that the curves cross each other so that the top and bottom switch. Hence we split the integral into two integrals:

[begin{align*} int_{-1}^{0}big[ 3(x^3-x)-0big] dx +int_{0}^{1}big[0-3(x^3-x) big] dx &= left(dfrac{3}{4}x^4-dfrac{3x^2}{2}right]_{-1}^0 - left(dfrac{3}{4}x^4-dfrac{3x^2}{2}right]_0^1 &=big(-dfrac{3}{4}+dfrac{3}{2} big) - big(dfrac{3}{4}-dfrac{3}{2} big) &=dfrac{3}{2} end{align*}.] Gmk dmg.

Application

Let (y = f(x)) be the demand function for a product and (y = g(x)) be the supply function. Then we define the equilibrium point to be the intersection of the two curves. The consumer surplus is defined by the area above the equilibrium value and below the demand curve, while the producer surplus is defined by the area below the equilibrium value and above the supply curve.

Example 4

Find the producer surplus for the demand curve

[ f(x) = 1000 - 0.4x^2 nonumber]

and the supply curve of

[ g(x) = 42x. nonumber]

Solution Hundreds boardanne 28 online, free games jewel quest.

Get on top clout games multiplayer. We first find the equilibrium point:

We set

[ 1000 - 0.4x^2 = 42x nonumber ]

or

7.5 Area Between Curves Ap Calculus Algebra

[ 0.4x^2 + 42x - 1000 = 0. nonumber]

We get

[x=20 nonumber]

hence

[y=42(20)=840. nonumber]

We integrate

[ begin{align*} int_{0}^{20} left ( 840 - 42x right ) dx &= {left[ 840x-21x^2 right] }_0^{20} [4pt] &= 8400. end{align*}]

7.5 Area Between Curves Ap Calculus 14th Edition

Exercises

1. Find the area between the curves ( y = x^2 ) and ( y =sqrt{x} ).
2. Find the area between the curves ( y = x^2 - 4) and ( y = -2x ).
3. Find the area between the curves ( y = 2/x ) and ( y = -x + 3 ).
4. Find the area between the curves ( y = x3^x ) and ( y = 2x +1 ).

Contributors and Attributions

• Larry Green (Lake Tahoe Community College)

• Integrated by Justin Marshall.