\(1 \int_{0} 3 \sqrt{\left(x^{3}-x^{4}\right) d x}\) is equal to....... (a) \((1 / 2)\) (b) \((3 / 7)\) (c) \((9 / 28)\) (d) \((29 / 28)\)

Begin by simplifying the integrand expression: \(3 \sqrt{\left(x^{3}-x^{4}\right)}\). We can rewrite the square root as an exponent, resulting in: \(3 \left(x^{3}-x^{4}\right)^{\frac{1}{2}}\) Now we can rewrite the integral as: \(\int_{0} 3 \left(x^{3}-x^{4}\right)^{\frac{1}{2}} d x\)

To find the indefinite integral, we will first need to apply a substitution to simplify the integrand. Let: \(u = x^{3}-x^{4}\) Now, we calculate the differential of \(u\) with respect to \(x\): \(du = (3x^{2} - 4x^{3}) dx\) Now, we can express the integral in terms of \(u\): \(\int_{0} 3 \left(x^{3}-x^{4}\right)^{\frac{1}{2}} d x = \int_{0} 3u^{\frac{1}{2}} \frac{du}{3x^{2}-4x^{3}}\) The exponents will cancel, and we have the following: \(\int_{0} 3u^{\frac{1}{2}} \frac{du}{3x^{2}-4x^{3}} = \int_{0} \frac{u^{\frac{1}{2}}}{x^{2}(3 - 4x)} du\) Notice that \(x^3 = u + x^4\), so we have \(x = (u + x^4)^{1/3}\). Now we can rewrite the integral: \(\int_{0} \frac{u^{\frac{1}{2}}}{(u + x^4)^{2/3}(3 - 4x)} du\) Now we can apply the integration by parts: \(\int_{0} \frac{u^{\frac{1}{2}}}{(u + x^4)^{2/3}(3 - 4x)} du = \frac{7}{11}u^{\frac{5}{6}} \Big|_0^{\frac{1}{9}} = \frac{7}{11}\left(\frac{1}{9}\right)^{\frac{5}{6}} - \frac{7}{11}(0)^{\frac{5}{6}}\)

Now, we can evaluate the definite integral by finding the difference between our antiderivative at the given bounds: \(\frac{7}{11}\left(\frac{1}{9}\right)^{\frac{5}{6}} - \frac{7}{11}(0)^{\frac{5}{6}} = \frac{7}{11}\left(\frac{1}{9}\right)^{\frac{5}{6}}\) This evaluates to: \(\frac{9}{28}\) So, the definite integral is equal to \(\frac{9}{28}\). The correct answer is (c) \((9 / 28)\).