Question

If $I_{1} = \int_{0}^{1} \frac{x^{5/2}(1-x)^{7/2}}{12} dx$; $I_{2} = \int_{0}^{1} \frac{x^{5/2}(1-x)^{7/2}}{(3+x)^{8}} dx$ then

• A. $I_{2} = 144\sqrt{3}I_{1}$
• B. $I_{1} = 864\sqrt{3}I_{2}$
• C. $I_{1} = I_{2}$
• D. $I_{1}I_{2} = 144\sqrt{3}$

Answer 1

Answer: (B)

$I_{1} = 864\sqrt{3}I_{2}$

Answer 2

Let $I_{1} = \int_{0}^{1} \frac{x^{5/2}(1-x)^{7/2}}{12} dx$ and $I_{2} = \int_{0}^{1} \frac{x^{5/2}(1-x)^{7/2}}{(3+x)^{8}} dx$.

First, let's evaluate $I_1$. The integral is related to the Beta function, $B(m, n) = \int_{0}^{1} x^{m-1}(1-x)^{n-1} dx = \frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}$. In the integral part of $I_1$, we have $x^{5/2}(1-x)^{7/2}$. Comparing with $x^{m-1}(1-x)^{n-1}$, we have $m-1 = 5/2 \implies m = 7/2$ and $n-1 = 7/2 \implies n = 9/2$. So, $\int_{0}^{1} x^{5/2}(1-x)^{7/2} dx = B(7/2, 9/2)$. $I_1 = \frac{1}{12} B(7/2, 9/2) = \frac{1}{12} \frac{\Gamma(7/2)\Gamma(9/2)}{\Gamma(7/2+9/2)} = \frac{1}{12} \frac{\Gamma(7/2)\Gamma(9/2)}{\Gamma(8)}$. Using the property $\Gamma(z+1) = z\Gamma(z)$ and $\Gamma(1/2) = \sqrt{\pi}$: $\Gamma(7/2) = \frac{5}{2}\Gamma(5/2) = \frac{5}{2} \cdot \frac{3}{2}\Gamma(3/2) = \frac{5}{2} \cdot \frac{3}{2} \cdot \frac{1}{2}\Gamma(1/2) = \frac{15}{8}\sqrt{\pi}$. $\Gamma(9/2) = \frac{7}{2}\Gamma(7/2) = \frac{7}{2} \cdot \frac{15}{8}\sqrt{\pi} = \frac{105}{16}\sqrt{\pi}$. $\Gamma(8) = 7! = 5040$. $B(7/2, 9/2) = \frac{(\frac{15}{8}\sqrt{\pi})(\frac{105}{16}\sqrt{\pi})}{5040} = \frac{\frac{15 \times 105}{8 \times 16}\pi}{5040} = \frac{1575\pi}{128 \times 5040}$. $\frac{1575}{5040} = \frac{225 \times 7}{720 \times 7} = \frac{225}{720} = \frac{45 \times 5}{144 \times 5} = \frac{45}{144} = \frac{5 \times 9}{16 \times 9} = \frac{5}{16}$. So, $B(7/2, 9/2) = \frac{5\pi}{128}$. $I_1 = \frac{1}{12} \cdot \frac{5\pi}{128} = \frac{5\pi}{1536}$.

Now, let's evaluate $I_2 = \int_{0}^{1} \frac{x^{5/2}(1-x)^{7/2}}{(3+x)^{8}} dx$. This integral is of the form $\int_{0}^{1} \frac{x^{a-1}(1-x)^{b-1}}{(c+dx)^{a+b}} dx$. Here, $a-1=5/2 \implies a=7/2$, $b-1=7/2 \implies b=9/2$. $a+b = 7/2 + 9/2 = 16/2 = 8$. The denominator is $(3+x)^8$, which matches $(c+dx)^{a+b}$ with $c=3$ and $d=1$. We use the integral formula: $\int_{0}^{1} \frac{x^{a-1}(1-x)^{b-1}}{(c+dx)^{a+b}} dx = \frac{1}{(c+d)^a c^b} B(a, b)$. In our case, $a=7/2$, $b=9/2$, $c=3$, $d=1$. $I_2 = \frac{1}{(3+1)^{7/2} 3^{9/2}} B(7/2, 9/2) = \frac{1}{4^{7/2} 3^{9/2}} B(7/2, 9/2)$. $4^{7/2} = (2^2)^{7/2} = 2^7 = 128$. $3^{9/2} = 3^4 \sqrt{3} = 81\sqrt{3}$. $I_2 = \frac{1}{128 \cdot 81\sqrt{3}} B(7/2, 9/2)$. We know $B(7/2, 9/2) = \frac{5\pi}{128}$. $I_2 = \frac{1}{128 \cdot 81\sqrt{3}} \cdot \frac{5\pi}{128} = \frac{5\pi}{128^2 \cdot 81\sqrt{3}} = \frac{5\pi}{16384 \cdot 81\sqrt{3}}$.

Now let's find the relationship between $I_1$ and $I_2$. $I_1 = \frac{1}{12} B(7/2, 9/2)$. $I_2 = \frac{1}{128 \cdot 81\sqrt{3}} B(7/2, 9/2)$. Let $B = B(7/2, 9/2)$. $I_1 = \frac{1}{12} B$. $I_2 = \frac{1}{128 \cdot 81\sqrt{3}} B$. Divide $I_1$ by $I_2$: $\frac{I_1}{I_2} = \frac{\frac{1}{12} B}{\frac{1}{128 \cdot 81\sqrt{3}} B} = \frac{1}{12} \cdot (128 \cdot 81\sqrt{3}) = \frac{128 \cdot 81\sqrt{3}}{12} = \frac{128 \cdot 27 \cdot 3\sqrt{3}}{4 \cdot 3} = 32 \cdot 27\sqrt{3} = 864\sqrt{3}$. So, $I_1 = 864\sqrt{3} I_2$.