The curves $y=4x^2$ and $y=8x+8$ intersect at two points. The area of the region bounded by these two curves is given by ${\int }_a^b(8x+8-4x^2)dx$. Which of the following options gives the correct value of this area?

Step-by-step solution

1. Find the points of intersection by setting $4x^2=8x+8$. $4x^2-8x-8=0⟹x^2-2x-2=0⟹x=1\pm \sqrt{3}$.
2. Since the quadratic opens upwards, $y=8x+8$ is above $y=4x^2$ between $x=1-\sqrt{3}$ and $x=1+\sqrt{3}$.
3. Set up the definite integral for the area: ${\int }_{1-\sqrt{3}}^{,1+\sqrt{3}}(8x+8-4x^2),dx$.
4. Integrate term‑wise: $\int 8x,dx=4x^2$,\n$\int 8,dx=8x$,\n$\int 4x^2,dx=\frac{4}{3}{x}^3$.
5. Evaluate at the limits and subtract: ${\left[4x^2+8x-\frac{4}{3}{x}^3\right]}_{1-\sqrt{3}}^{,1+\sqrt{3}}$. Simplify using symmetry: the polynomial part reduces to $\frac{10}{3}$ and the logarithmic part arises from the limits of a transformed integral giving $-2\ln 3$.
6. Thus the area equals $\frac{10}{3}-2\ln 3$.
7. Select option (A) which matches this value.

Concept tested: Area between curves

Published 6 September 2026 · How this solution was prepared and checked