Problem 1. $$ \int_{0}^{\infty} x^{7} e^{-x^{2}} \mathrm{~d} x $$ Problem 2. $$ \int_{0}^{\pi} \sin ^{5}(x) \mathrm{d} x $$

Question

UpStudy AI · Accepted Answer

**Problem 1**

We wish to evaluate
$$
\int_{0}^{\infty} x^{7} e^{-x^{2}} \, dx.
$$

1. Make the substitution $$\, u = x^2 $$. Then, $$\, du = 2x \, dx$$ which gives $$\, dx = \frac{du}{2x}$$.

2. Expressing $$x^7$$ as $$x^6 \cdot x$$ and using the substitution, we have:
   $$
   \int_{0}^{\infty} x^{7} e^{-x^2} \, dx 
   = \int_{0}^{\infty} x^{6} e^{-u} \cdot x\, \frac{du}{2x} 
   = \frac{1}{2}\int_{0}^{\infty} x^{6} e^{-u} \, du.
   $$

3. Since $$u = x^2$$, we have $$x^6 = (x^2)^3 = u^3$$. Thus, the integral becomes:
   $$
   \frac{1}{2}\int_{0}^{\infty} u^3 e^{-u} \, du.
   $$

4. Recall the Gamma function definition:
   $$
   \Gamma(n) = \int_{0}^{\infty} u^{n-1} e^{-u} \, du.
   $$
   Here, with $$n = 4$$, we have:
   $$
   \int_{0}^{\infty} u^3 e^{-u} \, du = \Gamma(4).
   $$

5. Knowing that $$\Gamma(4) = 3!\ = 6$$, we obtain:
   $$
   \int_{0}^{\infty} x^{7} e^{-x^2} \, dx = \frac{1}{2} \cdot 6 = 3.
   $$

**Problem 2**

We wish to evaluate
$$
\int_{0}^{\pi} \sin^5 (x) \, dx.
$$

1. Notice that the integrand $$\sin^5 x$$ is symmetric about $$x=\frac{\pi}{2}$$. Therefore, we can write:
   $$
   \int_{0}^{\pi} \sin^5 x \, dx = 2\int_{0}^{\pi/2} \sin^5 x \, dx.
   $$

2. One can use the Beta function representation for the integral:
   $$
   \int_{0}^{\pi/2} \sin^{m}(x) \, dx = \frac{\sqrt{\pi}\,\Gamma\left(\frac{m+1}{2}\right)}{2\,\Gamma\left(\frac{m}{2}+1\right)}.
   $$
   For $$m = 5$$, this becomes:
   $$
   \int_{0}^{\pi/2} \sin^5 x \, dx = \frac{\sqrt{\pi}\,\Gamma(3)}{2\,\Gamma(7/2)}.
   $$

3. Since $$\Gamma(3) = 2$$ and 
   $$
   \Gamma\left(\frac{7}{2}\right) = \frac{5}{2} \cdot \frac{3}{2} \cdot \frac{1}{2} \sqrt{\pi} = \frac{15\sqrt{\pi}}{8},
   $$
   we have:
   $$
   \int_{0}^{\pi/2} \sin^5 x \, dx = \frac{\sqrt{\pi}\cdot 2}{2\cdot \frac{15\sqrt{\pi}}{8}} = \frac{2\sqrt{\pi}}{\frac{15\sqrt{\pi}}{4}} = \frac{8}{15}.
   $$

4. Doubling to account for the interval $$[0,\pi]$$ gives:
   $$
   \int_{0}^{\pi} \sin^5 x \, dx = 2 \cdot \frac{8}{15} = \frac{16}{15}.
   $$

**Final Answers**

1. $$\displaystyle \int_{0}^{\infty} x^{7} e^{-x^{2}} \, dx = 3$$.

2. $$\displaystyle \int_{0}^{\pi} \sin^5(x) \, dx = \frac{16}{15}$$.
**Problem 1 Solution:** $$ \int_{0}^{\infty} x^{7} e^{-x^{2}} \, dx = 3 $$ **Problem 2 Solution:** $$ \int_{0}^{\pi} \sin^5(x) \, dx = \frac{16}{15} $$

Problem 1.
Problem 2.

Upstudy AI Solution

Answer

Solution

Beyond the Answer

Related Questions

Latest Calculus Questions

Problem 1. Problem 2.