If intlimits_0^1 {(4x^3 - f(x))f(x)dx = frac{ | vedclass

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(B) Given $\int\limits_0^1 {(4x^3 - f(x))f(x)dx = \frac{4}{7}}$.Expanding the integrand: $\int\limits_0^1 {(4x^3f(x) - (f(x))^2)dx = \frac{4}{7}}$.Rea

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$\frac{15}{2}$

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(B) Given $\int\limits_0^1 {(4x^3 - f(x))f(x)dx = \frac{4}{7}}$.Expanding the integrand: $\int\limits_0^1 {(4x^3f(x) - (f(x))^2)dx = \frac{4}{7}}$.Rearranging the terms: $\int\limits_0^1 {(-(f(x))^2 + 4x^3f(x))dx = \frac{4}{7}}$.Multiply by $-1$: $\int\limits_0^1 {((f(x))^2 - 4x^3f(x))dx = -\frac{4}{7}}$.Complete the square by adding $\int\limits_0^1 {(2x^3)^2 dx}$ to both sides:$\int\limits_0^1 {((f(x))^2 - 4x^3f(x) + 4x^6)dx = -\frac{4}{7} + \int\limits_0^1 {4x^6 dx}}$.$\int\limits_0^1 {(f(x) - 2x^3)^2 dx = -\frac{4}{7} + [\frac{4x^7}{7}]_0^1 = -\frac{4}{7} + \frac{4}{7} = 0}$.Since the integral of a squared function is $0$,we have $f(x) - 2x^3 = 0$,so $f(x) = 2x^3$.The area bounded by $y = 2x^3$,the $x$-axis,$x = 1$,and $x = 2$ is $\int\limits_1^2 {2x^3 dx}$.$= [\frac{2x^4}{4}]_1^2 = [\frac{x^4}{2}]_1^2 = \frac{16}{2} - \frac{1}{2} = \frac{15}{2}$.

If $\int\limits_0^1 {(4x^3 - f(x))f(x)dx = \frac{4}{7}}$,then the area of the region bounded by $y = f(x)$,the $x$-axis,and the ordinates $x = 1$ and $x = 2$ is:

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