Question

Find the area enclosed between the curves x2 = y and the straight line y = 3x+4​

Answer 1

Answer:

Answer

y=3x and y=x

2

insqunits

3x=x

2

x=3andx=0 are the points of intersection

0

∫

3

(3x−x

2

)dx

=

2

3x

2

∣

0

3

−

3

x

3

∣

0

3

=3

2

9

−

3

27

=

2

27

−9=

2

9

=4.5squnits

Answer 2

The area enclosed between the curves x² = y and the straight line y = 3x+4​ is $\frac{27}{2}$ sq. units.

Given:

x² = y

y = 3x+4​

To find:

The area enclosed between the curves x² = y and the straight line y = 3x+4​.

Solution:

x² = y                          .............................(1)

y = 3x+4​                     ..............................(2)

Put the value of y from equation 1 in equation 2.

x²= 3x+4​  

x²-3x-4=0

Use middle term splitting method and find the value of x.

​ x²-3x-4=0

x²-4x+x-4=0

x(x-4)-1(x-4)=0

(x-1)(x-4)=0

x= 1,4

We have to integrate x²-3x-4 for the interval 1 to 4 to find the area enclosed by the curves and the straight line.

Area enclosed between the curves and the straight line= $|\int\limits^4_1 {(x^{2} -3x-4)} \, dx|$

= $|[\frac{x^{3}}{3}-3\frac{x^{2}}{2}-4x]|^{4} _{1}$

= $|(\frac{64}{3}-\frac{1}{3})- 3(\frac{16}{2} -\frac{1}{2} )- 4(4-1) |$

=  $|(\frac{63}{3})- 3(\frac{15}{2} )- 4(3) |$

= $|21- (\frac{45}{2} )- 12 |$

=  $|9- (\frac{45}{2} ) |$

=  $|- (\frac{27}{2} ) |$

= $\frac{27}{2}$ sq. units

Therefore, the area enclosed between the curves x² = y and the straight line y = 3x+4​ is $\frac{27}{2}$ sq. units.

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