I need some help with various calculus problems! Please!

[alectronancy]

1) Evaluate the integral t^{23} lnt dt

2) Evaluate the integral from -2 to 1 of 1/{(x+7)(x^2+ 9 )} dx

3) Evaluate the indefinite integral sin^5(X) cos^8(X) dx

4) Evaluate the indefinite integral x^3 * square root (100+x^2) dx

5) Evaluate the indefinite integral ln(x)/x^4 dx

6) Consider the function f(x) = (x^2/3) + 8. Calculate using right endpoints, then find the limit as n approaches infinity.

Thanks for any help!

 

[soroban]

Hello, alectronancy!

Here's some help . . .

\(\displaystyle 1)\;\;I \;=\;\int t^{23}\ln t\,dt\)

\(\displaystyle \text{Integrate "by parts": }\;\begin{array}{ccccccc}u &=& \ln t & & dv &=& t^{23}dt \\ du &=& \frac{dt}{t} && v &=& \frac{1}{24}t^{24} \end{array}\)

\(\displaystyle \text{Then: }\;I \;=\;\frac{1}{24}\,t^{24}\ln t \:-\: \frac{1}{24}\!\int t^{23}dt \quad\hdots\;\text{etc.}\)

\(\displaystyle 2)\;\;\int_{-2}^1 \frac{dx}{(x+7)(x^2+ 9 )}\)

\(\displaystyle \text{Partial Fractions: }\;\frac{1}{(x+7)(x^2+9)} \;=\;\frac{A}{x+7} + \frac{Bx + C}{x^2+9}\)

\(\displaystyle \text{You should get: }\:A = \tfrac{1}{58},\;B = -\tfrac{1}{58},\;C = \tfrac{7}{58}\)

\(\displaystyle \text{Got it?}\)

\(\displaystyle 3)\;\;I \;=\;\int \sin^5\!x\cos^8\!x\,dx\)

\(\displaystyle \text{We have: }\;\sin^5\!x\cos^8\!x \;=\;\sin^4\!x\cos^8\!x\cdot \sin x\)

. . . . . . . . . . . . . . . .\(\displaystyle = \;(\sin^2\!x)^2\cos^8\!x\cdot\sin x\)

. . . . . . . . . . . . . . . .\(\displaystyle =\; (1-\cos^2\!x)^2\cos^8\!x\cdot\sin x\)

. . . . . . . . . . . . . . . .\(\displaystyle =\; (1 - 2\cos^2\!x + \cos^4\!x)\cos^8\!x\cdot\sin x\)

. . . . . . . . . . . . . . . .\(\displaystyle =\; (\cos^8\!x - 2\cos^{10}\!x + \cos^{12}\!x)\,\sin x\)


\(\displaystyle \text{Then: }\;I \;=\;\int(\cos^8\!x - 2\cos^{10}\!x + \cos^{12}\!x)\,\sin x\,dx\)

\(\displaystyle \text{Let: }\:u = \cos x \quad\Rightarrow\quad du = -\sin x\,dx \quad\Rightarrow\quad \sin x\,dx = -du\)

\(\displaystyle \text{Substitute: }\;I \;=\;\int(u^8 - 2u^{10} + u^{12})(-du) \;=\;-\int(u^8 - 2u^{10} + u^{12})\,du\)


\(\displaystyle \text{You can finish it, right?}\)

\(\displaystyle 4)\;\;I \;=\;\int x^3 \sqrt{100+x^2}\, dx\)

\(\displaystyle \text{Trig Substitution: }\;x \,=\,10\tan\theta \quad\Rightarrow\quad dx \,=\,10\sec^2\!\theta\,d\theta\)

. . . . . . . . . \(\displaystyle \text{and: }\;\sqrt{100+x^2} \,=\,10\sec\theta\)


\(\displaystyle \text{Substitute: }\;I \;=\;\int(10\tan\theta)^3\cdot10\sec\theta\cdot10\sec^2\!\theta\,d\theta\)

. . . . . . . . . .\(\displaystyle =\; 10^5\int \sec^3\!\theta\tan^3\!\theta\,d\theta\)

. . . . . . . . . .\(\displaystyle =\; 10^5\int\sec^2\theta\tan^2\theta(\sec\theta\tan\theta\,d\theta)\)

. . . . . . . . . .\(\displaystyle =\; 10^5\int\sec^2\theta(\sec^2\theta-1)(\sec\theta\tan\theta\,d\theta)\)

. . . . . . . . . .\(\displaystyle =\; 10^5\int(\sec^4\!\theta - \sec^2\!\theta)(\sec\theta\tan\theta\,d\theta)\)


\(\displaystyle \text{Let: }\;u \,=\,\sec\theta \quad\Rightarrow\quad du \,=\,\sec\theta\tan\theta\,d\theta\)

\(\displaystyle \text{Substitute: }\;I \;=\;10^5\int (u^4 - u^2)\,du\)


\(\displaystyle \text{Go for it!}\)

\(\displaystyle 5)\;\;I \;=\;\int\frac{\ln x}{x^4}\,dx\)

\(\displaystyle \text{By parts: }\;\begin{array}{ccccccc}u &=& \ln x && dv &=& x^{-4}dx \\ du &=& \frac{dx}{x} && v &=& -\frac{1}{3}x^{-3} \end{array}\)


\(\displaystyle I \;\;=\;\;-\tfrac{1}{3}x^{-3}\ln x \:+\: \tfrac{1}{3}\int x^{-4}dx \quad\hdots\;\text{etc.}\)

 

[galactus]

6) Consider the function f(x) = (x^2/3) + 8. Calculate using right endpoints, then find the limit as n approaches infinity.

What are the limits?. Were you given any?. Say, 0 to1 or something like that?.

Anyway, what we are doing with a Riemann sum is finding the area of an infinite number of rectangles. Thus, integrating and finding the area.

The right hand method means we use the right side of the rectangles and find their areas.

\(\displaystyle x_{k}=a+k{\Delta}x\). 'a' is your lower limit.

So, we have \(\displaystyle a+k\cdot \frac{b-a}{n}\)

You have, \(\displaystyle f(x)=\frac{x^{2}}{3}+8\).

The area of rectangle k is \(\displaystyle f(x_{k}){\Delta}x\)

\(\displaystyle \left(\frac{(a+k\cdot \frac{b-a}{n})^{2}}{3}+8\right)\frac{b-a}{n}\)

Where \(\displaystyle {\Delta}x=\frac{b-a}{n}\) is the length of each subinterval.

Think of \(\displaystyle {\Delta}x\) as the length of the rectangle and \(\displaystyle f(x_{k})\) as the height.

length times height is area of the rectangle. In this case, we sum all of those up.

Let me know the limits and I can help further if need be.