33.sagews

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2 (a)
\int_{2/3}^{\infty} \frac{dx}{9x^2+4} \\

\textrm{  First, do the following u-substitution}

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u = 3x, du = 3 dx, 

\textrm{ so we have the new integral }
\int \frac{du}{3(u^2+2^2)} = \frac{1}{6} \arctan ( \frac{3x}{2})

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\textrm{evaluating at endpoints we get}

\lim_{x\rightarrow \infty}  \frac{1}{6} \arctan ( \frac{3x}{2})- \frac{1}{6} \arctan ( 1)
=\frac{\pi}{12}-\frac{\pi}{24} = \frac{\pi}{24}

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2(b) \int_e^{e^2}\frac{dx}{x (\ln x)^p}

\textrm{ First, make the u-substitution }

u = \ln x, du = \frac{dx}{x}

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\textrm{this gives us the integral}
\int \frac{du}{u^p} 

\textrm{ which has two cases }
p =1, p \neq 1

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\textrm{If} p = 1,
\textrm{ we get }

\int_e^{e^2}\frac{dx}{x (\ln x)^p} = \ln \ln x |_{e}^{e^2} = \ln 2 - \ln 1

\textrm{If} p neq 1,\\

\int_e^{e^2}\frac{dx}{x (\ln x)^p} = \frac{(\ln x)^{1-p}}{p-1}|_{e}^{e^2} = \frac{2^{1-p}-1}{1-p}

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3(c) \int x3^{-2x} dx

\textrm{For this we must use integration by parts with }
dv = 3^{-2x}, u = x, v = \frac{3^{-2x}}{-2\ln 3}, du = dx

\textrm{note that } \int a^x dx \textrm{ where a is a constant is } \frac{a^x}{\ln a}

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\textrm{Thus we have } 

 \int x3^{-2x} dx = \frac{x3^{-2x}}{-2 \ln 3} - \frac{3^{-2x}}{-2 \ln 3} = \frac{1-x}{3^{2x}2 \ln 3}

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3 \textrm { The equation for the tangent line approximation at } a = 0 \textrm{ is }

f(x) = f(0)+f'(0)(x-0) = 1 -x/2

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\textrm{To make the two estimates, we just notice that they are }

\sqrt{0.9} = f(0.1) = 1-0.1/2 = 21/20 

\textrm{ and } 

\sqrt{0.99} = f(0.01) = 1 - 0.001/2 = 201/200

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4 (a) f(0) = 0

(b) f'(x) = \sqrt{4 - \sin(\pi t)} dt

(c) f(x) = f(0) - f'(0)(x-0) = 2x

(d) \textrm{For our error bound, we want } M \textrm{ such that }

\mid f"(x)\mid \leq M \textrm{ for all } x \textrm{ in the interval } [-1/6,1/6]

f"(x) = -\pi \cos( \pi x)/2 \sqrt{4 - \sin( \pi x)}

\textrm{For the next step, there are many ways to find such } M 

\textrm{We could use the first derivative text to see if the function is strictly increasing or decreasing and then use an endpoint.}

\textrm{Or we could use the second derivative text to find a maximum value on the interval.}

\textrm{However, the first and second derivatives are }

\pi^2\sin(\pi x)/(2\sqrt{4 - \sin(\pi x)}) - \pi^2\cos(\pi x)^2/(4(4 -
\sin(\pi x))^{3/2})

\textrm{and}

3\pi^3\cos(\pi x)\sin(\pi x)/(4(4 - \sin(\pi x))^{3/2}) +
\pi^3\cos(\pi x)/(2\sqrt{4 - \sin(\pi x)}) - 3\pi^3\cos(\pi x)^3/(8(4 -
\sin(\pi x))^(5/2))

\textrm{So it would be much easier to get a rough estimate by examing how large the numerator can be and how small the denominator can be.}  \cos (\pi x) \textrm{ reaches its maximum value at 0 and } \sin (\pi x) \textrm{ reaches its maximum value at the positive endpoint } 1/6.

\textrm{So the maximum the numerator can be is } \pi

\textrm{and the minimum the denominator can be is } \sqrt{4 - \sin(\pi/6)} = \sqrt{7/2}

\textrm{This gives us } M = \pi \sqrt{2/7}

(e) \textrm{We have the inequality } \mid error \mid \leq M(a-b)^2/2 \textrm{ where a and b are the endpoints of the interval where this inequality holds. Since we want our interval to be centered around zero, we can take } a = 0-x \textrm{ and } b = 0+x \textrm{ giving us the equation }

M(2x)^2/2 = 0.01  \textrm{ (Since we want the error less than 0.01) }

\textrm{Solving this equation for x we find the interval } 

[-\sqrt{0.01\sqrt{7}/(2\pi\sqrt{2})},\sqrt{0.01\sqrt{7}/(2\pi\sqrt{2})}]=[-0.0546,0.0546]