Question

If \[ \int (x+5)\sqrt{x-5}\,dx = \frac{2(x-5)^{5/2}}{15}f(x)+c, \] then \(f(6)=\)

• A. \(5\)
• B. \(20\)
• C. \(100\)
• D. \(53\)

Hint:

In substitution problems involving radicals like \(\sqrt{x-a}\), always use: \[ u=x-a \] This transforms the radical into a simple power of \(u\), making integration straightforward.

Answer 1

The Correct Option is D

Step 1: Substitute to remove the root.
Let $u=x-5$, so $x=u+5$ and $dx=du$. Then $x+5=u+10$.
Step 2: Rewrite the integral.
\[ \int(u+10)\sqrt{u}\,du=\int\left(u^{3/2}+10u^{1/2}\right)du. \]
Step 3: Integrate term by term.
\[ =\frac25u^{5/2}+10\cdot\frac23u^{3/2}+c=\frac25u^{5/2}+\frac{20}{3}u^{3/2}+c. \]
Step 4: Put $u=x-5$ back.
\[ =\frac25(x-5)^{5/2}+\frac{20}{3}(x-5)^{3/2}+c. \]
Step 5: Factor to match the given format.
Take out $\dfrac{2(x-5)^{3/2}}{15}$: \[ =\frac{2(x-5)^{3/2}}{15}\big[3(x-5)+50\big]+c=\frac{2(x-5)^{3/2}}{15}(3x+35)+c. \] So $f(x)=3x+35$.
Step 6: Evaluate $f(6)$.
$f(6)=3(6)+35=18+35=53$. \[ \boxed{53} \]