Question

Find the area bounded bounded by the following curves, y=x²/4 and x+4=2y.

Answer 1

Solution:

$y_1={x^2\over4}$

${2\over 2}y_2=x+4 \implies y_2={1\over 2}x+{4\over 2}\implies y_2={1\over 2}x+2$

At point of intersection, $y_1=y_2$

$\implies {x^2\over4}={1\over 2}x+2$

Multiply through by $8$

$\implies {x^2\over4}(8)={1\over 2}(8)x+2(8)$

$\implies 2x^2=4x+16$

Divide through by $2$

$\implies { 2\over 2}x^2={4\over 2}x+{16\over 2}$

$\implies x^2=2x+8 \implies x^2-2x-8 =0$

Solving quadratic equation:

$x^2-2x-8 =0$

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

$\implies x_1-4=0 \implies x_1=4$

$\implies x_2+2=0 \implies x_2=-2$

$Area=\int_{-2}^0(y_2-y_1)dx+\int_0^4(y_2-y_1)dx$

$=\int_{-2}^0({1\over 2}x+2-{x^2\over4})dx+\int_0^4({1\over 2}x+2-{x^2\over4})dx$

$=[{x^2\over 2\times 2}+2x-{x^3\over 4\times 3}]_{-2}^0+[{x^2\over 2\times 2}+2x-{x^3\over 4\times 3}]_0^4$

$=[{x^2\over 4}+2x-{x^3\over 12}]_{-2}^0+[{x^2\over 4}+2x-{x^3\over 12}]_0^4$

$=({0^2\over 4}+2(0)-{0^3\over 12})-({(-2)^2\over 4}+2(-2)-{(-2)^3\over 12})+({4^2\over 4}+2(4)-{4^3\over 12})-({0^2\over 4}+2(0)-{(0)^3\over 12})$

$=(0+0-0)-(1-4+0.666666666)+(4+8-5.333333333)-(0+0-0)$

$=4-1-0.666666666+4+8-5.333333333$

$=9.000000001$ square units