Question

(a) Evaluate the integral ∫2048x2+4dx. Your answer should be in the form kπ, where k is an integer. What is the value of k? (Hint: darctan(x)dx=1x2+1 ) k= 6 (b) Now, lets evaluate the same integral using power series. First, find the power series for the function f(x)=48x2+4. Then, integrate it from 0 to 2, and call it S. S should be an infinite series ∑[infinity]n=0an . What are the first few terms of S?

Answer 1

Answer:

Step-by-step explanation:

a) Evaluate the integral

$\int\limits^2_0 {\frac{48}{x^2+4} } \, dx$

$=\int\limits^2_0 {\frac{48}{x^2+2} } \, dx \\\\=\frac{48}{2} [\tan ^-^1(\frac{\pi}{2}) ]^2_0$

$=24[\tan ^-^1(1)- \tan^-^1(0)]\\\\=24(\frac{\pi}{4} )\\\\=6\pi\\\\ k \pi=6 \pi\\\\k=6$

b)

$F(x)=\frac{48}{x^2+4}$

divide by 4

$=\frac{12}{1+(\frac{x}{2} )^2}$

Power series for $\frac{1}{1-x}$

$\frac{1}{1-x} =1+x+x^2+x^3+x^4+...= \sum_{n=0}^{\infty}x^n |x|<1...(1)$

replace x by $-\frac{x^2}{4}$ in equation (1)

$\frac{12}{1+(\frac{x}{2})^2 } =12\frac{1}{1-(-\frac{x^2}{4} )}$

$=12 \sum_{n=0}^{\infty}(-\frac{x^2}{4} )^n\\\\=12\sum_{n=0}^ \infty(-1)^n\frac{x^2^n}{2^2^n}$

$=12(1-\frac{x^2}{4} +\frac{x^4}{16} -\frac{x^6}{64} +\frac{x^8}{256} -\frac{x^1^0}{1024} ...)$

$=12-3x^2+\frac{3}{4} x^4-\frac{3}{16} x^6+\frac{3}{64}x^8-\frac{3}{256} x^1^0...$

Take integration with respect to x from 0 to 2

$\int\limits^2_0 {(12-3x^2+\frac{3}{4}x^4-\frac{3}{16} x^6+\frac{3}{64} x^8-\frac{3}{256} x^1^0+... )} \, dx$

$=[{12x-x^3+\frac{3}{20} x^5-\frac{3}{16}\frac{x^7}{7} +\frac{3}{64} \frac{x^9}{9} -\frac{3}{256}\frac{x^1^1}{11} +...]^2_0 }$

$=24-8+4.8-3.42+2.66-2.18\\\\=18.84$