10/8/2021

11.1 Area Between Curvesap Calculus

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

  1. 11.1 Area Between Curves Ap Calculus Formulas
  2. 11.1 Area Between Curves Ap Calculus Algebra
  3. 11.1 Area Between Curves Ap Calculus Calculator
  4. 11.1 Area Between Curves Ap Calculus 14th Edition

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

By integrating the difference of two functions, you can find the area between them. Created by Sal Khan.Practice this lesson yourself on KhanAcademy.org righ.

  • The area under a general curve, you need calculus. Each of these situations involves the same general strategy—the reformulation of precalculus mathematics through the use of a limit process. So, one way to answer the question “What is calculus?” is to say that calculus is a “limit machine” that involves three stages.
  • We start by finding the area between two curves that are functions of ( displaystyle x ), beginning with the simple case in which one function value is always greater than the other. We then look at cases when the graphs of the functions cross. Last, we consider how to calculate the area between two curves that are functions of ( displaystyle.
  • 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)).

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]

To remember this formula we write

[ 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*}]

Algebra

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:

[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

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*}.]

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

We first find the equilibrium point:

We set

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

or

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

We get

[x=20 nonumber]

hence

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

11.1 Area Between Curves Ap Calculus Formulas

We integrate

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

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 ).

11.1 Area Between Curves Ap Calculus Algebra

Contributors and Attributions

  • Larry Green (Lake Tahoe Community College)
  • Integrated by Justin Marshall.

Area between Curves

The area between curves is given by the formulas below.

11.1 Area Between Curves Ap Calculus Calculator

Area = (int_a^b {,left {fleft( x right) - gleft( x right)} right ,dx} )

for a region bounded above by y = f(x) and below by y = g(x), and on the left and right by x = a and x = b.

for a region bounded on the left by x = f(y) and on the right by x = g(y), and above and below by y = c and y = d.

Example 1:1

Find the area between y = x and y = x2 from x = 0 to x = 1.

(eqalign{{rm{Area}} &= int_0^1 {left {x - {x^2}} right dx} &= int_0^1 {left( {x - {x^2}} right)dx} &= left. {left( {frac{1}{2}{x^2} - frac{1}{3}{x^3}} right)} right _0^1 &= left( {frac{1}{2} - frac{1}{3}} right) - left( {0 - 0} right) &= frac{1}{6}})

1

Find the area between x = y + 3 and x = y2 from y = –1 to y = 1.

(eqalign{{rm{Area}} &= int_{ - 1}^1 {left {y + 3 - {y^2}} right dy} &= int_{ - 1}^1 {left( {y + 3 - {y^2}} right)dy} &= left. {left( {frac{1}{2}{y^2} + 3y - frac{1}{3}{x^3}} right)} right _{ - 1}^1 &= left( {frac{1}{2} + 3 - frac{1}{3}} right) - left( {frac{1}{2} - 3 + frac{1}{3}} right) &= frac{{16}}{3}})

11.1 Area Between Curves Ap Calculus 14th Edition

See also

Area under a curve, definite integral, absolute value rules