Question

You are finding the area of the region bounded by y = f(x), the x-axis, and the lines x = 1 and x = 2.

What's given in the problem

• The integral $\int_0^1 (4x^3 - f(x))f(x) \,dx = -\frac{4}{7}$.

Answer 1

The area of the region bounded by $y = f(x)$, the x-axis, and the lines $x = 1$ and $x = 2$ can be $\frac{15(1+\sqrt{2})}{2}$ or $\frac{15(\sqrt{2}-1)}{2}$.

Answer 2

1. Rewrite the given integral $\int_0^1 (4x^3 - f(x))f(x) \,dx = -\frac{4}{7}$ by completing the square for $f(x)$.

2. The integrand $4x^3 f(x) - (f(x))^2$ becomes $4x^6 - (f(x) - 2x^3)^2$.

3. Substitute this back into the integral: $\int_0^1 (4x^6 - (f(x) - 2x^3)^2) \,dx = -\frac{4}{7}$.

4. Evaluate $\int_0^1 4x^6 \,dx = \frac{4}{7}$.

5. This leads to $\frac{4}{7} - \int_0^1 (f(x) - 2x^3)^2 \,dx = -\frac{4}{7}$, which simplifies to $\int_0^1 (f(x) - 2x^3)^2 \,dx = \frac{8}{7}$.

6. This condition does not uniquely determine $f(x)$. Assuming $f(x)$ is of the form $cx^3$, substitute it into the original integral to find possible values for $c$.

7. Solving $c^2 - 4c - 4 = 0$ yields $c = 2 \pm 2\sqrt{2}$.

8. Both $f(x) = (2+2\sqrt{2})x^3$ and $f(x) = (2-2\sqrt{2})x^3$ satisfy the derived integral condition.

9. Calculate the area $\int_1^2 |f(x)| \,dx$ for both functions.

   • For $f(x) = (2+2\sqrt{2})x^3$, area is $\frac{15(1+\sqrt{2})}{2}$.
   • For $f(x) = (2-2\sqrt{2})x^3$, area is $\frac{15(\sqrt{2}-1)}{2}$.

10. Since both functions satisfy the given condition, both areas are valid.