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Mathematics LibreTexts

Test 3

  • Page ID
    13715
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    Exercise \(\PageIndex{1}\)

    Calculate the following integrals:

    \[\displaystyle \int \frac{ x^2+1} {\sqrt{x}} dx\]

     \[\displaystyle \int_{1}^{2} (3x^2+2x-1) \,dx\]

    Answer

    \( \dfrac{2 x^{5/2}}{5}+ 2x^{1/2}+C\), \(9\)

     

    Exercise \(\PageIndex{2}\)

    Verify that the hypotheses of the Mean Value Theorem are satisfied on the given interval and find all values of \(c\) in that interval that satisfy the conclusion of the theorem for \(f(x) = x^2-x \) on \([-3,5].\)

    Answer

    \(c=1\)

    Solution:

    Since \(f \) is a polynomial, it is continuous and differentiable everywhere. Therefore, \(f\) satisfies the hypotheses of the Mean Value Theorem, and there must exist at least one value \( c∈(-3,5) \) such that \(f'(c)\) is equal to the slope of the secant line connecting \((-3,f(-3))\) and \((5,f(5)). \)

    To determine which value(s) of \(c\) are guaranteed, first calculate the derivative of \(f\). The derivative \(f′(x)= 2x-1\). The slope of the line connecting \((-3,f(-3))\) and \((5,f(5))\) is given by

    \[\frac{f(5)−f(-3)}{5+3}=\frac{20−12}{8}=\frac{8}{8}=1.\]

    We want to find \(c\) such that \(f′(c)=\frac{7}{4}\). That is, we want to find \(c\) such that

    \[2c-1=1\]

    Solving this equation for \(c\), we obtain \(c=1\).

    Exercise \(\PageIndex{3}\)

    Calculate the following integrals by using substitution:

    1. \(\displaystyle{\int \left( \frac{1}{\sqrt{t}} - 3\sqrt{t} \right) \, dt}\).
    2. \(\displaystyle{\int 4x\sec^2(x^2)\, dx}\).
    3. \(\displaystyle{\int \sin^4(x)\cos(x) \, dx}\).
    4. \(\displaystyle{\int {1 \over 5x+2} \, dx}\).
    5. \(\displaystyle{\int {\frac{dx}{\sqrt{x}(x+1)}}}\).
    Hint:

    2. use \(u= x^2\),

    3. use \(u= \sin(x) \),

    4. use \(u=5 x+2\),

    5. use \(u= \sqrt{x}\).

    Answer

    1. \(2t^{\frac{1}{2}}-2t^{\frac{3}{2}} +C\),

    2. \(2 \tan(x^2)+C\),

    3. \(\dfrac {\sin^5(x)}{5}+C\),

    4. \(\dfrac {\ln(5x+2)}{5}+C\),

    5. \(2 \tan^{-1}(\sqrt{x})+C \).

    Exercise \(\PageIndex{4}\)

    An oil tanker is leaking at the rate given (in barrels per hour) by \[L^{'}(t)= \displaystyle \frac{50 \ln(t+1)}{t+1},\] where $t$ is the time (in hours) (when $t=0$). Find the total number of barrels that ship will leak on the end of the first day.

    Answer

    \(25 ( ln(25))^2) \).

    Exercise \(\PageIndex{5}\)

    Solve the initial value problem \[ \frac{dy}{dt} = e^{-2t}, y(0)=5.\]

    Answer

    \( y= \dfrac {-e^{-2t}+11}{2}\).

    Exercise \(\PageIndex{6}\)

    Find the position, velocity, speed and acceleration at time \(t=1\), \[ s(t)=9-9\cos\left( \displaystyle \frac{\pi t}{3}\right), 0 \leq t \leq 5.\]

    Answer

    \(s(1)=4.5 ft. v(1) =1.5 \sqrt{3} \pi, a(1)= \pi^2/2.\)

    Solution:

    \(s(1)=9-9\cos\left( \displaystyle \frac{\pi }{3}\right)= 9-9(\dfrac{1}{2})=4.5 ft\)

    \(v(t)=s'(t)= 3 \pi \sin\left( \displaystyle \frac{\pi t}{3}\right)\), \(v(1)=s'(1)=3 \pi \sin\left( \displaystyle \frac{\pi }{3}\right)= 1.5 \sqrt{3} \pi\)

    \(a(t)=v'(t)= \pi ^2 \cos\left( \displaystyle \frac{\pi t}{3}\right)\), \(a(1)=v'(1)=\pi^2 \cos\left( \displaystyle \frac{\pi }{3}\right)=)= \pi^2/2\)

     

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