Question

Evaluate the integral ∫2032x2+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= 4 (b) Now, lets evaluate the same integral using power series. First, find the power series for the function f(x)=32x2+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

Here is the correct computation of the question;

Evaluate the integral :

`\int\limits^2_0 \ (32)/(x^2 +4) \ dx`

Your answer should be in the form kπ, where k is an integer. What is the value of k?

(Hint:
`(d \ arc \ tan (x))/(dx) =(1)/(x^2 + 1)`)

k = 4

(b) Now, lets evaluate the same integral using power series.

`f(x) = (32)/(x^2 +4)`

Then, integrate it from 0 to 2, and call it S. S should be an infinite series

What are the first few terms of S?

Answer:

(a) The value of k = 4

(b)

`a_0 = 16\\ \\ a_1 = -4 \\ \\ a_2 = (12)/(5) \\ \\a_3 = - (12)/(7) \\ \\ a_4 = (12)/(9)`

Explanation:

(a)

`\int\limits^2_0 (32)/(x^2 + 4) \ dx`

`= 32 \int\limits^2_0 (1)/(x+4)\ dx`

`=32 ((1)/(2) \ arctan ((x)/(2)))^2__0`

`= 32 ( (1)/(2) arctan ((2)/(2))- (1)/(2) arctan ((0)/(2)))`

`= 32 ( (1)/(2)arctan (1) - (1)/(2) arctan (0))`

`= 32 ( (1)/(2)((\pi)/(4))- (1)/(2)(0))`

`= 32 ((\pi)/(8)-0)`

`= 32 ( ((\pi)/(8)))`

`= 4 \pi`

The value of k = 4

(b)
`(32)/(x^2+4)= 8 - (3x^2)/(2^1)+ (3x^4)/(2^3)- (3x^6)/(2x^5)+ (3x^8)/(2^7) -... \ \ \ \ \ (Taylor\ \ Series)`

`\int\limits^2_0 (32)/(x^2+4)= \int\limits^2_0 (8 - (3x^2)/(2^1)+ (3x^4)/(2^3)- (3x^6)/(2x^5)+ (3x^8)/(2^7) -...) dx`

`S = 8 \int\limits^2_0dx - (3)/(2^1) \int\limits^2_0 x^2 dx + (3)/(2^3)\int\limits^2_0 x^4 dx - (3)/(2^5)\int\limits^2_0 x^6 dx+ (3)/(2^7)\int\limits^2_0 x^8 dx-...`

`S = 8(x)^2_0 - (3)/(2^1*3)(x^3)^2_0 +(3)/(2^3*5)(x^5)^2_0- (3)/(2^5*7)(x^7)^2_0+ (3)/(2^7*9)(x^9)^2_0-...`

`S= 8(2-0)-(1)/(2^1)(2^3-0^3)+(3)/(2^3*5)(2^5-0^5)- (3)/(2^5*7)(2^7-0^7)+(3)/(2^7*9)(2^9-0^9)-...`

`S= 8(2-0)-(1)/(2^1)(2^3)+(3)/(2^3*5)(2^5)- (3)/(2^5*7)(2^7)+(3)/(2^7*9)(2^9)-...`

`S = 16-2^2+(3)/(5)(2^2) -(3)/(7)(2^2) + (3)/(9)(2^2) -...`

`S = 16-4 + (12)/(5)- (12)/(7)+ (12)/(9)-...`

`a_0 = 16\\ \\ a_1 = -4 \\ \\ a_2 = (12)/(5) \\ \\a_3 = - (12)/(7) \\ \\ a_4 = (12)/(9)`