June 18th, 2014, 02:30 PM  #1 
Newbie Joined: Jun 2014 From: London Posts: 13 Thanks: 0  Integration.
Could any one possibly help me these questions. I have written the question and then my attempt at the solution next. Find each of the following indefinite integrals, identifying any general rules of calculus that you use. [integral]xcos(1/3x)dx Solving the equation by integration by parts. Let f(x)=x and g'(x)=cos(1/3x); then f'(x)=1 and g(x)=3 sin(1/3x) Substituting into the equation for integration by parts f(x)g(x)∫f ' (x)g(x)dx. we obtain ∫xcos(1/3x)dx=3xsin(1/3x)∫1(3 sin(1/3)x)=3xsin(1/3x)3∫sin(1/3)x=3xsin(1/3 x)+9cos(1/3 x)+C x/√1x^4 Take u=1x^4; thendu/dx=4x^3. Hence∫x/√(1x^4) dx=x/√u (4x^3 )dx=x/√u du= 1/2 u^(3/2). Plugging u back into the equation1/2 4x^3 ^(3/2) Explain why the graph of the function f(x) = (x2 + 1) ln(1/2x) lies below the xaxis for 1 < x < 2 and above the xaxis for 2 < x < 3. The graph of the function f(x)=(x^2+1) ln(1/2x) is positive above the xaxis and negative below the xaxis. Use this fact to find the area enclosed by the graph and the xaxis between x = 1 and x = 3, giving your answer to five decimal places. I have no idea how to answer this. Find the volume of the solid of revolution obtained when the graph of f(x)= sec x + tan x, from x=−[pie]/3 to x = [pie]/4 is rotated about the xaxis. Give your answer to four decimal places. The volume is [pie]∫([pie]/3) ([pie]/4) secx+tanx=[pie](ln(tan(x)+sec(x) )+ln(sec(x)))=[pie]([(ln(tan([pie]/3)+sec([pie]/3)+ln(sec([pie]/3) ][(ln(tan([pie]/4)+sec([pie]/4)+ln(sec([pie]/4)]=[pie [ln(√3+2)+ln(2)[(ln(1+√2)+ln(√2)]=[pie][0.631.2)=1.83[pie] 
June 18th, 2014, 02:51 PM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra 
In the first problem you have an error integrating your $g^\prime(x)$. In the second, try $\sin u = x^2$. In the third you can integrate by parts so that you idfferentiate the logarithmic term. The point about the function being negative is that the integral will be negative on that part of the interval. I think they want you to split the interval into two parts and integrate $f(x)$ on the negative part to get a positive result. 

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