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Area Between Two Curves

Find the area between two curves, triangle using calculus & integrating with respect to y. Practice problems on area calculations.

Area Between Two Given Curves

  • The area A of the region bounded by curves \(y=f(x),\;y=g(x)\) and the lines \(x=a,\,x=b\) where \(f,\,g\) are continuous and \(f(x)\geq g(x)\;\;\forall\,x\,\varepsilon[a,\,b]\) is 

\(\displaystyle A=\int\limits^b_a(f(x)-g(x))\,dx\) ...(1) 

  • If \(g(x)=0\) then the area becomes [equation of x axis is \(y=0\) ]

\(\displaystyle A=\int\limits^b_af(x)\,dx\) ...(2)

  • Sketch the region, identify the top and bottom curve.

Illustration Questions

Find the area of region bounded by the curve \(y=x^2+2,\;y=x\) and lines \(x=0\) and \(x=3\).

A 2

B 5

C \(\dfrac{17}{2}\)

D \(\dfrac{21}{2}\)

×

Sketch \(y=x^2+2\to\) It can be obtained from \(y=x^2\) by shifting vertically by two units.

image

Now draw the lines \(y=x,\;x=0\) and \(x=3\).

image

\(\displaystyle Area=\int\limits^b_a(f(x)-g(x))\,dx\)

\(f(x)=x^2+2,\;a=0\)

\(g(x)=x,\;b=3\)

\(\displaystyle A=\int\limits^3_0(x^2+2-x)\,dx\)

\(=\Bigg[\dfrac{x^3}{3}+2x-\dfrac{x^2}{2}\Bigg]^3_0\)

\(=\left(9+6-\dfrac{9}{2}\right)-(0)\)

\(=\dfrac{21}{2}\)

Find the area of region bounded by the curve \(y=x^2+2,\;y=x\) and lines \(x=0\) and \(x=3\).

A

2

.

B

5

C

\(\dfrac{17}{2}\)

D

\(\dfrac{21}{2}\)

Option D is Correct

Area Bounded by Two Curves which Intersects at Two Points

Consider two curves \(y=f(x)\) and \(y=g(x)\) as shown.

They intersect at points \(A\) and \(B\). The area bounded by them (shaded area) is given by

\(\displaystyle A=\int\limits^b_a(f(x)-g(x))\,dx\)

Where \('a'\) is the \(x\)-coordinate of point \(A\) and \('b'\) is the \(x\)-coordinate of point \(B\).

  • Lower limit = \(x\)- coordinate of extreme left point of area to be found.
  • Upper limit = \(x\)- coordinate of extreme right point of area to be found.

Illustration Questions

Find the area of shaded region given below.

A \(\dfrac{1}{2}\)

B \(\dfrac{3}{4}\)

C \(\dfrac{5}{6}\)

D \(\dfrac{9}{8}\)

×

Area \(\displaystyle=A=\int\limits^b_a(f(x)-g(x))\,dx\)

here \(f(x)=\dfrac{x+2}{4},\;g(x)=\dfrac{x^2}{4}\)

\(a=-1,\;b=2\)

\(\therefore\displaystyle\,A=\int\limits^2_{-1}\left(\dfrac{x+2}{4}-\dfrac{x^2}{4}\right)\,dx\)

\(\displaystyle=\dfrac{1}{4}\,\int\limits^2_{-1}(x+2-x^2)\,dx\)

\(=\dfrac{1}{4}\,\left[\dfrac{x^2}{2}+2x-\dfrac{x^3}{3}\right]^{2}_{-1}\)

\(=\dfrac{1}{4}\,\left[\left(\dfrac{4}{2}+4-\dfrac{8}{3}\right)-\left(\dfrac{1}{2}-2+\dfrac{1}{3}\right)\right]\)

\(=\dfrac{1}{4}\,\left[\dfrac{10}{3}+\dfrac{7}{6}\right]\)

\(=\dfrac{1}{4}×\dfrac{27}{6}\)

\(=\dfrac{9}{8}\)

Find the area of shaded region given below.

image
A

\(\dfrac{1}{2}\)

.

B

\(\dfrac{3}{4}\)

C

\(\dfrac{5}{6}\)

D

\(\dfrac{9}{8}\)

Option D is Correct

Area of Triangle using Calculus

  • Consider the area of triangle whose vertices are \(A(x_1,\,y_1),\;B(x_2,\,y_2)\) and \(C(x_3,\,y_3)\).

This area can be calculated by using the formula from coordinate geometry.

  • This area can also be evaluated by integration.

\(\displaystyle A=\int\limits^{x_2}_{x_1}(\ell_1(x)-\ell_2(x))\,dx+\int\limits^{x_3}_{x_2}(\ell_3(x)-\ell_2(x))\,dx\)

Where \(y=\ell_1(x),\;\ell_2(x)\) and \(\ell_3(x)\) an the equation of straight line \(AB,\,AC\) and \(BC\).

Illustration Questions

Find the area of region of triangle with vertices \((-1,\,0),\;(1,\,3)\) and \((3,\,2)\).

A 6

B 4

C \(\dfrac{5}{2}\)

D \(\dfrac{9}{4}\)

×

The extreme left point \(A\) has \(x\)- coordinate \(=-1\)

The extreme right point \(C\) has \(x\)- coordinate \(=3\)

image

The lower curve is the line \(AC\) in the interval \((-1,3) \).

The upper curve is \(AB\) in the interval \((-1,\,1)\) and \(BC\) in the interval\((1,\,3)\).

image

We need equation of all sides of the triangle

Equation of \(AB\to y-0=\dfrac{3-0}{1-(-1)}(x-(-1))\)

\(\Rightarrow\,y=\dfrac{3}{2}(x+1)\)

\(\Rightarrow\,2y=3x+3\)

\(\Rightarrow\,3x-2y+3=0\)

Equation of \(BC\to y-2=\dfrac{3-2}{1-3}(x-3)\)

\(\Rightarrow\,y-2=-\dfrac{1}{2}(x-3)\)

\(\Rightarrow\,2y-4=-x+3\)

\(\Rightarrow\,x+2y-7=0\)

Equation of \(AC\to y-0=\dfrac{2-0}{3-(-1)}(x-(-1))\)

\(\Rightarrow\,y=\dfrac{1}{2}(x+1)\)

\(\Rightarrow\,2y=x+1\)

\(\Rightarrow\,x-2y+1=0\)

image

Now Area \(\displaystyle=A=\int\limits^1_{-1}\underbrace{\left(y_{AB}-y_{AC}\right)}_{A_1}\,dx+\int\limits^3_{1}\underbrace{\left(y_{BC}-y_{AC}\right)}_{A_2}\,dx\)

image

\(A_1=\displaystyle\,\int\limits^1_{-1}\left(\dfrac{3x+3}{2}-\dfrac{x+1}{2}\right)\,dx\)

\(=\displaystyle\int\limits^1_{-1}(x+1)\,dx\)

\(=\Bigg[\dfrac{x^2}{2}+x\,\Bigg]^1_{-1}\)

\(=\left(\dfrac{1}{2}+1\right)-\left(\dfrac{1}{2}-1\right)\)

\(=2\)

image

\(A_2=\displaystyle\,\int\limits^3_{1}\left(\dfrac{7-x}{2}-\dfrac{x+1}{2}\right)\,dx\)

\(=\displaystyle\int\limits^3_{1}\,3-x\,dx\)

\(=\Bigg[3x-\dfrac{x^2}{2}\,\Bigg]^3_{1}\)

\(=\left(9-\dfrac{9}{2}\right)-\left(3-\dfrac{1}{2}\right)\)

\(=\dfrac{9}{2}-\dfrac{5}{2}\)

\(=2\)

image

\(A=A_1+A_2=4\)

image

Find the area of region of triangle with vertices \((-1,\,0),\;(1,\,3)\) and \((3,\,2)\).

A

6

.

B

4

C

\(\dfrac{5}{2}\)

D

\(\dfrac{9}{4}\)

Option B is Correct

Area by Integrating with Respect to "y"

  • Some regions can be treated by expressing \(x\) as a function of \(y\). The area bounded by \(x=f(y),\;x=g(y)\) and lines \(y=c,\;y=d\) is given by 

\(\displaystyle\,=\int\limits^d_c(f(y)-g(y))\,dy\)

  • \(f(y)=x\) is the curve to the right and \(x=g(y)\) to the left.   (i.e. \(f(y)\geq g(y)\)).
  • Observe the shape of area to be found and decide upon the variable to the take \((x\;or\;y)\).

Illustration Questions

Find the area enclosed by the line \(x-y-1=0\) and the parabola \(y^2=2x+1\).

A \(\dfrac{16}{3}\)

B 4

C \(\dfrac{5}{2}\)

D \(\dfrac{12}{5}\)

×

Intersection point of \(y=x-1\) and \(y^2=2x+1\) are obtained by solving both equations.

\(\therefore\,(x-1)^2=2x+1\)

\(\Rightarrow\,x^2-2x+1=2x+1\)

\(\Rightarrow\,x^2-4x=0\)

\(\Rightarrow\,x(x-4)=0\)

\(\Rightarrow\,x=0,\;4\)

\(\therefore\) points are \((0,-1)\) and \((4,\,3)\)

image

Area \(\displaystyle=A=\int\limits^d_c(f(y)-g(y))\,dy\)

Here \({c=-1}\,\,\,\,\,(y \,\,coordinate \,\,of\,\, extreme\,\, bottom\,\, point\,\, of\,\, area)\) , 

\({d=3}\)      \( (y\,\, coordinate\,\, at\,\, extreme\,lower\, point\,\, of\,\, the \,\,curve)\),

\(f(y)=y+1,\;g(y)=\dfrac{y^2-1}{2}\)

image

\(\displaystyle=A=\int\limits^3_{-1}\left[(y+1)-\left(\dfrac{y^2-1}{2}\right)\right]\,dy\)

\(\displaystyle=\int\limits^3_{-1}\left(y-\dfrac{y^2}{2}+\dfrac{3}{2}\right)\,dy\)

\(=\Bigg[\dfrac{y^2}{2}-\dfrac{y^3}{6}+\dfrac{3}{2}y\Bigg]^3_{-1}\)

\(=\left(\dfrac{9}{2}-\dfrac{9}{2}+\dfrac{9}{2}\right)-\left(\dfrac{1}{2}+\dfrac{1}{6}-\dfrac{3}{2}\right)\)

\(=\dfrac{9}{2}-\left(\dfrac{-5}{6}\right)\)

\(=\dfrac{32}{6}\)

\(=\dfrac{16}{3}\)

image

Find the area enclosed by the line \(x-y-1=0\) and the parabola \(y^2=2x+1\).

A

\(\dfrac{16}{3}\)

.

B

4

C

\(\dfrac{5}{2}\)

D

\(\dfrac{12}{5}\)

Option A is Correct

Area Bounded by two Parabolas

If the area bounded by two parabolas, one of the form \(y^2=4\,ax\) and other of the form \(x^2=4\,by\) is asked then sketch both parabolas and apply the formula

\(A=\displaystyle\int\limits^b_a(f(x)-g(x))\,dx\)

Illustration Questions

Find the area bounded by parabolas \(y^2=4x\) and \(x^2=4y\).

A \(\dfrac{32}{3}\)

B \(\dfrac{16}{3}\)

C 7

D 6

×

\(Area=A=\displaystyle\int\limits^b_a(f(x)-g(x))\,dx\)  \(\to\;f(x)=2\sqrt x,\;g(x)=\dfrac{x^2}{4}\)

image

Solve for intersection points of \(y^2=4x\) and \(x^2=4y\)

We get \(\left(\dfrac{x^2}{4}\right)^2=4x\)

\(\Rightarrow\,\dfrac{x^4}{16}=4x\)

\(\Rightarrow\,x^3=64\)

\(\Rightarrow\,x=4,\,0\)

\(\therefore\,B\) is \((4,\,4)\),  \(A\) is \((0,\,0)\)

image

\(A=\displaystyle\int\limits^4_0\left(2\sqrt x-\dfrac{x^2}{4}\right)\,dx\)

\(=\Bigg[\dfrac{2x^{3/2}}{\dfrac{3}{2}}-\dfrac{x^3}{12}\Bigg]^4_0\)

\(=\left(\dfrac{4}{3}×4^{3/2}-\dfrac{64}{12}\right)-0\)

\(=\dfrac{32}{3}-\dfrac{16}{3}\)

\(=\dfrac{16}{3}\)

image

Find the area bounded by parabolas \(y^2=4x\) and \(x^2=4y\).

A

\(\dfrac{32}{3}\)

.

B

\(\dfrac{16}{3}\)

C

7

D

6

Option B is Correct

Calculation of Area

  • To find the area bounded by two curves we first sketch both the curves and identify the area to be found.
  • Then find the \(x\) coordinate of extreme left point of the required area. This will be lower lower limit of area integral = \(a\).
  • Find the \(x\)- coordinate of extreme right point of required area. This will be upper limit of area integral = \(b\).

 

Illustration Questions

Find the area bounded by parabolas \(y=5x^2\) and \(y=2x^2+9\).

A \(4\sqrt 2\)

B \(12\sqrt 3\)

C 15

D 6

×

First, we sketch the graph of \(y=2x^2+9\)

image

The graph of \(y=5x^2\)

image

The combined graph of both the curves,

image

Solving for intersection points of \(y=5x^2\) and \(y=2x^2+9\)

we get \(5x^2=2x^2+9\)

\(\Rightarrow\,x=\pm\sqrt5\)

\(\therefore\,A\) and \(B\) are \((-\sqrt3,\;15)\) and \((\sqrt3,\,15)\)

\(A=Area=\displaystyle\int\limits^b_a(f(x)-g(x))\,dx\)

where \(a=-\sqrt 3,\;b=\sqrt 3\)

\(f(x)=2x^2+9,\;g(x)=5x^2\)

\(A=\displaystyle\int\limits^{\sqrt 3}_{-\sqrt 3}(2x^2+9)-(5x^2)\,dx\)

\(=\displaystyle\int\limits^{\sqrt 3}_{-\sqrt 3}(9-3x^2)\,dx\)

\(=\Bigg[9x-x^3\Bigg]^{\sqrt3}_{-\sqrt3}\)

\(=(9\sqrt3-3\sqrt3)-(-9\sqrt3+3\sqrt 3)\)

\(=18\sqrt3-6\sqrt3\)

\(=12\sqrt3\)

Find the area bounded by parabolas \(y=5x^2\) and \(y=2x^2+9\).

A

\(4\sqrt 2\)

.

B

\(12\sqrt 3\)

C

15

D

6

Option B is Correct

Calculation of Area when two Curves Cross Each Other

  • The upper curve should be the one from which lower one is subtracted, taking modulus ensure this

Shaded Area = \(\displaystyle\int\limits^b_a\left|f(x)-g(x)\right|\,dx\)

\(\displaystyle\int\limits^c_a\underbrace{\left(f(x)-g(x)\right)}_{A_1}\,dx+\displaystyle\int\limits^b_c\underbrace{\left(g(x)-f(x)\right)}_{A_2}\,dx\)

\(|x|=x\) if \(x\geq 0\)

\(=-x\) if \(x<0\)

Therefore

\(|f(x)-g(x)|=f(x)-g(x)\) if \(f(x)>g(x)\) and  \(|f(x)-g(x)|=g(x)-f(x)\) if \(f(x)<g(x)\)

 

Illustration Questions

Find the area bounded by the curves \(y=x^3\) and the line \(y=x\).

A 2

B \(\dfrac{1}{2}\)

C \(\dfrac{5}{4}\)

D \(\dfrac{4}{5}\)

×

Sketch the graph first

image

\(A=Area=\displaystyle\int\limits^b_a\left|x-x^3\right|\,dx\)

where \('a'\) and \('b'\) are \(x\) -coordinate of extreme point of intersects.

image

Solving for point of intersection of \(y=x^3\) and \(y=x\)

we get \(x^3=x\)

\(\Rightarrow\,x=0,\;-1,\,1\)

\(\therefore\,D(-1,\,-1),\;B(0,\,0)\) and \(C(1,1)\) are points of intersect.

image

\(A=\displaystyle\int\limits^1_{-1}\left|x-x^3\right|\,dx\)

\(=\displaystyle\int\limits^0_{-1}\left(x^3-x\right)\,dx+\int\limits^1_0x-x^3\,dx\)

\(=\Bigg[\dfrac{x^4}{4}-\dfrac{x^2}{2}\Bigg]^0_{-1}+\Bigg[\dfrac{x^2}{2}-\dfrac{x^4}{4}\Bigg]^1_{0}\)

\(=\left[0-\left(\dfrac{1}{4}-\dfrac{1}{2}\right)\right]+\left(\dfrac{1}{2}-\dfrac{1}{4}\right)\)

\(=\dfrac{1}{4}+\dfrac{1}{4}\)

\(=\dfrac{1}{2}\)

 

image

Find the area bounded by the curves \(y=x^3\) and the line \(y=x\).

A

2

.

B

\(\dfrac{1}{2}\)

C

\(\dfrac{5}{4}\)

D

\(\dfrac{4}{5}\)

Option B is Correct

Finding the Values of Parameters when certain Relations and Area Values are given

  • Sometimes the area bounded by two curves (often are of them is a line) is given and one of the curves certain a parameter.
  • To find the value of this parameter we find the area which is a function of this parameter and equation it to the area value given.

Equation of the area bounded by \(y=x^2\) and \(y=\alpha x\;(\alpha>0)\) \(x\) given by

\(Area=\displaystyle\int\limits^{\alpha}_{0}\left(\alpha x-x^2\right)\,dx\) which depends on \(\alpha\).

If the value of the area is given, \(\alpha\) can be found.

Illustration Questions

If the area bounded by the parabola \(y^2=9x\) and the line \(y=mx\) is \(\dfrac{27}{2}\), find the value of \(m\).

A \(m=1\)

B \(m=2\)

C \(m=-1\)

D \(m=4\)

×

Sketch \(y^2=9x\) and the line \(y=mx\).

image

Solve for intersection point of \(y^2=9x\) and \(y=mx\) 

we get

\(m^2x^2=9x\)

\(x=0,\;9/m^2\)

\(\therefore\,B\left(\dfrac{9}{m^2},\,\dfrac{9}{m}\right)\) and \(O(0,\,0)\) are the points of intersection.

image

\(Area=A=\displaystyle\int\limits^b_a\left(f(x)-g(x)\right)\,dx\)

Here \(a=0,\;b=\dfrac{9}{m^2},\;f(x)=3\sqrt x,\;g(x)=mx\)

image

\(\therefore\,\dfrac{27}{2}=\displaystyle\int\limits^{9/m^2}_{0}\left(3\sqrt x-mx\right)\,dx\)

\(\Rightarrow\,\dfrac{27}{2}=\Bigg[\dfrac{3x^{3/2}}{\dfrac{3}{2}}-\dfrac{mx^2}{2}\Bigg]^{9/m^2}_0\)

\(\dfrac{27}{2}=2×\left(\dfrac{9}{m^2}\right)^{3/2}-m×\left(\dfrac{9}{m^2}\right)^2×\dfrac{1}{2}\)

\(\Rightarrow\,\dfrac{27}{2}=\dfrac{54}{m^3}-\dfrac{81}{2m^3}\)

\(\Rightarrow\,\dfrac{1}{2}=\dfrac{2}{m^3}-\dfrac{3}{2m^3}\)

\(\Rightarrow\,\dfrac{1}{2}=\dfrac{1}{2m^3}\)

\(\Rightarrow\,m^3=1\)

\(\Rightarrow\,m=1\)

image

If the area bounded by the parabola \(y^2=9x\) and the line \(y=mx\) is \(\dfrac{27}{2}\), find the value of \(m\).

A

\(m=1\)

.

B

\(m=2\)

C

\(m=-1\)

D

\(m=4\)

Option A is Correct

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