Question

Find the area of region y =17x²/16, y = 9x²/4 and y = 1 - 4x, satisfying x ≤0.

Answer 1

$$\dfrac{17x^2}{16}=1-4x, x\le0$$

$$17x^2+64x-16=0$$

$$(17x-4)(x+4)=0$$

Since $x\le0,$ we take $x=-4.$

$$\dfrac{9x^2}{4}=1-4x, x\le0$$

$$9x^2+16x-4=0$$

$$(9x-2)(x+2)=0$$

Since $x\le0,$ we take $x=-2.$

$$\dfrac{17x^2}{16}=\dfrac{9x^2}{4}$$

$$x=0$$

$$A=\displaystyle\int_{-4}^{-2}(1-4x-\dfrac{17x^2}{16})dx$$

$$+\displaystyle\int_{-2}^{0}(\dfrac{9x^2}{4}-\dfrac{17x^2}{16})dx$$

$$=[x-2x^2-\dfrac{17x^3}{48}]\begin{matrix} -2\\ -4 \end{matrix}+[\dfrac{19x^3}{48}]\begin{matrix} 0\\ -2 \end{matrix}$$

$$=-2-8+\dfrac{17}{6}+4+32-\dfrac{68}{3}+\dfrac{19}{6}$$

$$=\dfrac{28}{3}({units}^2)$$

Area= $\dfrac{28}{3}$ square units.