Informative line

### 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.

#### 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.

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

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

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

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

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

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

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

$$=\dfrac{21}{2}$$

### 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.

#### 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.

A

$$\dfrac{1}{2}$$

.

B

$$\dfrac{3}{4}$$

C

$$\dfrac{5}{6}$$

D

$$\dfrac{9}{8}$$

Option D 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

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

#### 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

×

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

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 $$(0,\,0)$$

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

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

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

$$=\dfrac{32}{3}-\dfrac{16}{3}$$

$$=\dfrac{16}{3}$$

### 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 = $$b$$.

#### 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 $$y=2x^2+9$$

The graph of $$y=5x^2$$

The combined graph of both the curves,

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

# 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 $$BC$$.

#### 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 $$=3$$

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$$(1,\,3)$$.

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

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

$$\Rightarrow\,2y=x+1$$

$$\Rightarrow\,x-2y+1=0$$

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

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

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

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

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

$$=2$$

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

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

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

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

$$=\dfrac{9}{2}-\dfrac{5}{2}$$

$$=2$$

$$A=A_1+A_2=4$$

### 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 $$(x\;or\;y)$$.

#### 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 $$(4,\,3)$$

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

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

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

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

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

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

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

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

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

$$=\dfrac{32}{6}$$

$$=\dfrac{16}{3}$$

### 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

# 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 $$f(x)<g(x)$$

#### 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

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

where $$'a'$$ and $$'b'$$ are $$x$$ -coordinate of extreme point of intersects.

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.

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

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

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

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

$$=\dfrac{1}{4}+\dfrac{1}{4}$$

$$=\dfrac{1}{2}$$

### 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.

#### 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 $$y=mx$$.

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.

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

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

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

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

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

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

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

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

$$\Rightarrow\,m^3=1$$

$$\Rightarrow\,m=1$$

### 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