Question

Find the area bounded by the curves x² = 4y and the straight line x - 4y + 2 = 0.

Answer 1

x-4y+2=0
y=x-2y+2y+2
y=x-2y+2
y=x-2(y)
y=x-2y
y=4y-2y
y=2y

Answer 2

Answer:

$\frac{9}{8}$ Sq. units.

Step-by-step explanation:

We have to calculate the area bounded by the curve x²=4y ........ (1)

and the straight line x-4y+2=0, ⇒y=$\frac{x}{4}+\frac{1}{2}$ ........ (2)

Now, solving equations (1) and (2), we get

x²=4($\frac{x}{4}+\frac{1}{2}$)

⇒$x^{2} -x-2=0$

⇒(x-2)(x+1) =0

⇒ x=2 or -1

Hence, the curve and the straight line meet with each others at x=2 and at x=-1.

Hence, the required area will be {From equations (1) and (2)}

∫₋₁²[($\frac{x}{4}+\frac{1}{2}$)-$\frac{x^{2} }{4}$]dx

=[$\frac{x^{2} }{8}+\frac{x}{2}-\frac{x^{3} }{12}$]₋₁²

=$\frac{4}{8}+\frac{2}{2}-\frac{8}{12}-\frac{1}{8}+\frac{1}{2}-\frac{1}{12}$

=$\frac{12+24-16-3+12-2}{24}$

=$\frac{27}{24}$

=$\frac{9}{8}$ Sq. units. (Answer)