Given that \displaystyle \int_0^4 x^3\sqrt{9+x^2} dx = a, what is the value of \lfloor a \rfloor?
I get 14
To find the value of \(\lfloor a \rfloor\), we need to evaluate the definite integral \(\displaystyle \int_{0}^{4} x^{3} \sqrt{9+x^{2}} \, dx\) and find the value of \(a\).
To evaluate this integral, we can use a substitution. Let's set \(u = 9+x^{2}\), which implies \(du = 2x \, dx\). Now, we need to express \(x\) and \(dx\) in terms of \(u\) and \(du\).
Rearranging the equation \(u = 9 + x^{2}\), we have \(x^{2} = u - 9\), which means \(x = \sqrt{u-9}\).
Substituting \(dx\) with \(\frac{du}{2x}\), we have \(dx = \frac{du}{2\sqrt{u-9}}\).
Now, we can rewrite the integral with the substitution:
\(\displaystyle \int_{0}^{4} x^{3} \sqrt{9+x^{2}} \, dx = \int_{9}^{25} \left(\sqrt{u-9}\right)^{3} \frac{du}{2\sqrt{u-9}}\).
Simplifying the expression inside the integral, we have:
\(\displaystyle \int_{9}^{25} \frac{1}{2} (u-9)^{\frac{3}{2}} \, du\).
To evaluate this integral, we can use the power rule of integration. Applying the power rule, we get:
\(\displaystyle \frac{1}{2} \cdot \frac{2}{5} \cdot (u-9)^{\frac{5}{2}} \Bigg|_{9}^{25}\).
Simplifying further, we have:
\(\displaystyle \frac{1}{5} \left[(25-9)^{\frac{5}{2}} - (9-9)^{\frac{5}{2}}\right]\).
\(\displaystyle \frac{1}{5} \left[16^{\frac{5}{2}} - 0^{\frac{5}{2}}\right]\).
Since \(0^{\frac{5}{2}}\) is equal to zero, the expression becomes:
\(\displaystyle \frac{1}{5} \cdot 16^{\frac{5}{2}}\).
Evaluating \(16^{\frac{5}{2}}\) gives us:
\(\displaystyle \frac{1}{5} \cdot 256\).
Finally, computing the value of \(\frac{1}{5} \cdot 256\) gives us the value of \(a\).
Thus, the value of \(\lfloor a \rfloor\) is the greatest integer less than or equal to the value of \(a\).