Question:

Step 1: Recognize the Beta function form in $I_1$. We begin by identifying that $I_1$ can be expressed using the Beta function. We have: $$I_1 = \frac{1}{30}\int_0^1 x^{7/2}(1-x)^{9/2}\,dx$$ This integral matches the Beta function form $B(p,q) = \int_0^1 x^{p-1}(1-x)^{q-1}\,dx$ with $p = 9/2$ and $q = 11/2$. Therefore: $$I_1 = \frac{B(9/2, 11/2)}{30}$$ Step 2: Apply the Beta function formula to $I_2$. For $I_2$, we use the generalized Beta integral formula: $$\int_0^1 \frac{x^{p-1}(1-x)^{q-1}}{(x+c)^{p+q}}\,dx = \frac{B(p,q)}{c^q(1+c)^p}$$ With $p = 9/2$, $q = 11/2$, and $c = 5$, we get: $$I_2 = \int_0^1 \frac{x^{7/2}(1-x)^{9/2}}{(x+5)^{10}}\,dx = \frac{B(9/2, 11/2)}{5^{11/2} \cdot 6^{9/2}}$$ Step 3: Calculate the ratio $\frac{I_1}{I_2}$. Now we compute the ratio of the two integrals: $$\frac{I_1}{I_2} = \frac{\dfrac{B(9/2, 11/2)}{30}}{\dfrac{B(9/2, 11/2)}{5^{11/2} \cdot 6^{9/2}}}$$ The Beta functions cancel: $$\frac{I_1}{I_2} = \frac{5^{11/2} \cdot 6^{9/2}}{30}$$ Step 4: Simplify the ratio using exponent rules. Since $30 = 5 \cdot 6$, we can write: $$\frac{I_1}{I_2} = \frac{5^{11/2} \cdot 6^{9/2}}{5 \cdot 6} = 5^{11/2 - 1} \cdot 6^{9/2 - 1} = 5^{9/2} \cdot 6^{7/2}$$ Step 5: Express the result in the required form. We can rewrite this as: $$5^{9/2} \cdot 6^{7/2} = 5 \cdot 5^{7/2} \cdot 6^{7/2} = 5 \cdot (5 \cdot 6)^{7/2} = 5 \cdot 30^{7/2}$$ Step 6: Match with the given form $5a^3\sqrt{a}$. We are given that $\frac{I_1}{I_2} = 5a^3\sqrt{a}$, which can be rewritten as: $$5a^3\sqrt{a} = 5a^{7/2}$$ From Step 5, we have: $$5 \cdot 30^{7/2} = 5a^{7/2}$$ Dividing both sides by 5: $$30^{7/2} = a^{7/2}$$ Therefore: $$a = 30$$ **Final Answer:** The value of $a$ is $\boxed{30}$, which corresponds to **Option 4**.