Calculus Early Transcendentals: Integral & Multi-Variable Calculus for Social Sciences

Find the following indefinite integrals.

1. \(\ds\int 8\sqrt{x}\,dx\)

   Answer
   \(\ds(16/3)x^{3/2}+C\)
   Solution

   An antiderivative of \(f(x) = \sqrt{x}\) is \(F(x) = \frac{2}{3} x^{3/2}\text{.}\) Therefore,

   \begin{equation*} \int 8\sqrt{x} \,dx = 8\int \sqrt{x}\,dx = 8\left(\frac{2}{3} x^{3/2} + C_1\right) = \frac{16}{3} x^{3/2} + C\text{,} \end{equation*}

   for some constant \(C\text{.}\)

2. \(\ds\int (3t^2+1) \,dt\)

   Answer
   \(\ds t^3+t+C\)
   Solution

   We use the indefinite integral rules:

   \begin{equation*} \begin{split} \int 3t^2+1\,dt \amp = 3\int t^2\,dt + \int 1 \,dt \\ \amp = 3 \left(\frac{1}{3}t^3+C_1\right) + \left(t+C_2\right)\\ \amp = t^3 + t + C. \end{split} \end{equation*}

3. \(\ds\int \frac{4}{\sqrt{y}} \,dy\)

   Answer
   \(\ds 8\sqrt{y}+C\)
   Solution

   We use the indefinite integral rules:

   \begin{equation*} \begin{split} \int \frac{4}{\sqrt{y}}\,dy \amp = 4\int y^{-1/2}\,dy \\ \amp = 4\left(2\sqrt{y}+C_1\right)\\ \amp = 8\sqrt{y}+C \end{split} \end{equation*}

4. \(\ds\int \frac{2}{z^2} \, dz\)

   Answer
   \(-2/z+C\)
   Solution

   We use the indefinite integral rules:

   \begin{equation*} \begin{split} \int \frac{2}{z^2}\,dy \amp = 2\int z^{-2}\,dz \\ \amp = 2\left(-z^{-1}+C_1\right)\\ \amp = -\frac{2}{z} + C \end{split} \end{equation*}

5. \(\ds\int 7s^{-1}\,ds\)

   Answer
   \(7 \ln s + C\)
   Solution

   We use the indefinite integral rules:

   \begin{equation*} \begin{split} \int 7s^{-1}\,ds \amp = 7\int s^{-1}\,ds \\ \amp = 7\left(\ln |s| + C_1\right)\\ \amp = 7\ln|s|+C \end{split} \end{equation*}

6. \(\ds\int (5x+1)^2\,dx\)

   Answer
   \(\ds (5x+1)^3/15+C\)
   Solution

   We use the indefinite integral rules:

   \begin{equation*} \begin{split} \int \left(5x+1\right)^2 \,dx \amp = \int \left(25x^2+10x+1\right)\,dx \\ \amp = 25\int x^2\,dx + 10\int x\,dx + \int 1\,dx \\ \amp = 25 \left(\frac{1}{3} x^3 + C_1\right) + 10\left(\frac{1}{2}x^2 + C_2\right) + x + C_3\\ \amp = \frac{25}{3} x^3 + 5x^2 + x + C \end{split} \end{equation*}

7. \(\ds\int (t-6)^2 \,dt\)

   Answer
   \(\ds (t-6)^3/3+C\)
   Solution

   We use the indefinite integral rules:

   \begin{equation*} \begin{split} \int (t-6)^2\,dt \amp = \int \left(t^2-12t+36\right)\,dt\\ \amp = \int t^2\,dt -12\int t\,dt + 36\int\,dt \\ \amp = \left(\frac{1}{3}t^3+C_1\right) - 12\left(\frac{1}{2}t^2+C_2\right) + 36\left(t+C_3\right)\\ \amp = \frac{1}{3}t^3 - 6t^2 + 36t + C \end{split} \end{equation*}

8. \(\ds\int z^{3/2}\,dz\)

   Answer
   \(\ds 2z^{5/2}/5+C\)
   Solution

   We use the indefinite integral rules:

   \begin{equation*} \begin{split} \int z^{3/2}\,dz \amp = \frac{2}{5} z^{5/2} + C \end{split} \end{equation*}

9. \(\ds\int {2\over x\sqrt x} \, dx\)

   Answer
   \(\ds -4/\sqrt{x}+C\)
   Solution

   We use the indefinite integral rules:

   \begin{equation*} \begin{split} \int \frac{2}{x\sqrt{x}}\,dx \amp = 2\int x^{-3/2}\,dx \\ \amp = 2\left(-2x^{-1/2}+C_1\right)\\ \amp = -\frac{4}{\sqrt{x}}+ C \end{split} \end{equation*}

10. \(\ds\int|2t-4| \, dt\)

    Answer
    \(\ds 4t-t^2+C\text{,}\) \(t\lt 2\text{;}\) \(\ds t^2-4t+8+C\text{,}\) \(t\ge 2\)
    Solution

    We first rewrite the integrand using the definition of the absolute value function:

    \begin{equation*} |2t-4| = \begin{cases}2t-4 \amp t \geq 2\\ 4-2t \amp t \lt 2 \end{cases} \end{equation*}

    We first consider the case where \(t \geq 2\text{:}\)

    \begin{equation*} \begin{split} \int 2t-4\,dt \amp = 2\int t\,dt - 4\int \,dt \\ \amp = 2\left(\frac{1}{2}t^2+C_1\right) - 4\left(t+C_2\right)\\ \amp = t^2 -4t + C \end{split} \end{equation*}

    Similarly, when \(t \lt 2\text{,}\) we have

    \begin{equation*} \begin{split} \int 4-2t\,dt \amp = -\int 2t-4\,dt \\ \amp = -t^2 + 4t + C \end{split} \end{equation*}

    All together, we have

    \begin{equation*} \int |2t-4|\,dt = \begin{cases}t^2 -4t + C \amp t \geq 2\\ 4t -t^2 + C \amp t \lt 2 \end{cases} \end{equation*}

11. \(\ds\int \left(\cos(2x)+4\sin(x)\right)\,dx\)

    Answer
    \(\ds (\sin x-4)\cdot \cos x + C\)
    Solution

    By linearity,

    \begin{equation*} \int \left(\cos(2x)+4\sin(x)\right)\,dx = \int \cos(2x) \,dx + 4 \int \sin(x)\,dx\text{.} \end{equation*}

    We know that \(\diff{}{x}\cos(x) = -\sin(x)\text{.}\) Therefore,

    \begin{equation*} \int \sin(x)\,dx = -\cos(x) + C_1\text{.} \end{equation*}

    Now, by the chain rule,

    \begin{equation*} \diff{}{x} \sin(2x)= 2\cos(2x) \implies \frac{1}{2}\diff{}{x} \sin(2x)= \cos(2x)\text{,} \end{equation*}

    and so

    \begin{equation*} \int \cos(2x) \,dx = \frac{1}{2} \sin(2x) + C_2\text{.} \end{equation*}

    Together, we have

    \begin{equation*} \int \left(\cos(2x)+4\sin(x)\right)\,dx = \frac{1}{2}\sin(2x)-4\cos(x)+C\text{,} \end{equation*}

    for some constant \(C\text{.}\)

12. \(\ds\int \left(2y-\sec^2y\right)\,dy\)

    Answer
    \(\ds y^2 - \tan y + C\)
    Solution

    We use the indefinite integral rules:

    \begin{equation*} \begin{split} \int (2y - \sec^2 y) \,dy \amp = 2\int y\,dy - \int \sec^2 y\,dy\\ \amp = 2\left(\frac{1}{2}y^2+C_1\right) - \left(\tan y + C_2\right)\\ \amp = y^2 - \tan y + C \end{split} \end{equation*}

13. \(\ds\int \left(\frac{\tan x}{\cos x} - \csc^2 x\right) \,dx\)

    Answer
    \(\ds \cot x + \sec x + C\)
    Solution

    We use the indefinite integral rules:

    \begin{equation*} \begin{split} \int \left(\frac{\tan x}{\cos x} - \csc^2 x\right)\,dx \amp = \int \frac{\sin x}{\cos^2 x}\,dx - \int \csc^2x \,dx \\ \amp = \left(\sec x+ C_1\right) - \left(-\cot x + C_2\right)\\ \amp = \sec x + \cot x + C \end{split} \end{equation*}

Find the following indefinite integrals. Hint

1. \(\ds \int \frac{2s^3+s^2-s}{s} ds\)

   Answer
   \(\ds \frac{s}{6}\left(4s^2+3s-6\right)+C\)
   Solution

   \(\begin{aligned}\displaystyle{\int\frac{2s^3+s^2-s}{s}\,ds} \amp = \int\left(2s^2+s-1\right)\,ds \\[1ex] \amp = \frac{2}{3}s^3 +\frac{1}{2}s^2-s + C \end{aligned}\)

2. \(\ds \int \frac{3x^3-x^2+x-2}{x} dx\)

   Answer
   \(\ds x^3 - \frac{1}{2}x^2 + x - 2\ln|x| + C\)
   Solution

   \(\begin{aligned}\displaystyle{\int \frac{3x^3-x^2+x-2}{x}\,dx} \amp = \int \left(3x^2-x+1-\frac{2}{x}\right)\,dx \\ \amp = x^3 - \frac{1}{2}x^2 + x - 2\ln|x| + C \end{aligned}\)

3. \(\ds \int \frac{u^3 + \sqrt[3]{2u}}{2u^2} du\)

   Answer
   \(\ds \frac{u^{8/3}-3\sqrt[3]{2}}{4u^{2/3}} + C\)
   Solution

   \(\begin{aligned}\int\frac{u^3+\sqrt[3]{2u}}{2u^2}\,du \amp =\int \left(\frac{1}{2}u + \frac{1}{2^{2/3}}u^{-5/3}\right)\,du \\[1ex] \amp = \frac{1}{4}u^2 + \frac{3}{2\cdot 2^{2/3}}u^{-2/3}+C \end{aligned}\)

4. \(\ds \int \frac{\left(\sqrt{t}-2\right)^2}{4t^2} dt\)

   Answer
   \(\ds \frac{8\sqrt{t}+t\ln t-4}{4t} + C\)
   Solution

   \(\begin{aligned}\int \frac{\left(\sqrt{t}-2\right)^2}{4t^2}\,dt \amp = \int \frac{t-4\sqrt{t}+4}{4t^2}\,dt \\ \amp = \int \left(\frac{1}{4}t^{-1} - t^{-3/2} + t^{-2}\right)\,dt\\ \amp = \frac{1}{4}\ln|t| +\frac{2}{\sqrt{t}}-\frac{1}{t} + C \end{aligned}\)

Answer the initial value problems to find \(f(t)\text{.}\)

1. \(f'(t)=3t+2\text{,}\) where \(f(0)=5\)

   Answer
   \(f(t) = \frac{3}{2}t^2+2t+5\)
   Solution

   If \(f'(t)=3t+2\text{,}\) then we must have

   \begin{equation*} f(t) = \int \left(3t+2\right)\,dt = \frac{3}{2}t^2 + 2t + C\text{.} \end{equation*}

   Therefore, if \(f(0) = 5\text{,}\) then we must have \(C=5\text{.}\) Hence,

   \begin{equation*} f(t) = \frac{3}{2}t^2+2t+5\text{.} \end{equation*}

2. \(f'(t)=\frac{t+3}{t}\text{,}\) where \(f(1)=-1\)

   Answer
   \(f(t)= t + 3\ln t - 2\)
   Solution

   We first integrate:

   \begin{equation*} \int f'(t)\,dt = \int\frac{t+3}{t}\,dt = \int \left(1+\frac{3}{t}\right)\,dt = t -3\ln|t| + C\text{.} \end{equation*}

   Since we further know that \(f(1)=-1\text{,}\) we can solve for the arbitrary constant \(C\text{:}\)

   \begin{equation*} f(1) = 1- 3\ln(1) + C = 1+C = -1 \implies C = -2\text{.} \end{equation*}

   Therefore, the solution to the initial value problem is

   \begin{equation*} f(t) = t - 3\ln|t| -2\text{.} \end{equation*}

3. \(f'(t)=\sin t\text{,}\) where \(f(\pi) = 0\)

   Answer
   \(f(t) = -\cos t-1\)
   Solution

   Since \(f'(t) = \sin(t)\text{,}\) the function \(f(t)\) must be of the form

   \begin{equation*} f(t) = \int \sin t\,dt = -\cos(t)+C\text{.} \end{equation*}

   To solve for the constant \(C\text{,}\) we set

   \begin{equation*} f(\pi) = 0 \implies 1+C = 0 \implies C = -1\text{.} \end{equation*}

   All together, we find that

   \begin{equation*} f(t) = -\cos(t) -1\text{.} \end{equation*}

4. \(f'(t) = e^{2t}-4\text{,}\) where \(f(0) = 1\)

   Answer
   \(f(t)=0.5e^{2t}-4t+0.5\)
   Solution

   We first integrate:

   \begin{equation*} \int e^{2t} - 4\,dt = \frac{1}{2}e^{2t} - 4t + C\text{.} \end{equation*}

   Therefore, if \(f(0) = 1\text{,}\) then

   \begin{equation*} f(0) = \frac{1}{2} + C = 1 \implies C = \frac{1}{2}\text{.} \end{equation*}

   Therefore, the solution to the initial value problem is

   \begin{equation*} f(t) = \frac{1}{2} e^{2t} - 4t + \frac{1}{2}\text{.} \end{equation*}

A retailer determines that the marginal revenue function associated with selling \(q\) items is

dollars per week per item.

1. Determine \(R(q)\text{.}\)

   Answer
   \(R(q) = 16q - 0.004q^2\)
   Solution

   We integrate the marginal revenue function:

   \begin{equation*} R(q) = \int \left(-0.008q + 16\right)\,dq = -0.004q^2+16q + C\text{.} \end{equation*}

   Since we must have that \(R(0)=0\) (i.e. there is no revenue if no units are sold), it follows that \(C=0\text{:}\)

   \begin{equation*} R(q) = -0.004q^2+16q\text{.} \end{equation*}

2. Determine the demand equation relating the unit price \(p\) to the quantity \(q\) demanded.

   Answer
   \(p(q) = \frac{16q - 0.004q^2}{q}\)
   Solution

   The demand equation \(p(q)\) must satisfy \(R(q) = p \times q\text{.}\) Therefore,

   \begin{equation*} p(q) = \frac{R(q)}{q} = -0.004q + 16\text{.} \end{equation*}

A supermarket determines that the marginal profit function associated with purchasing and selling \(q\) heads of lettuce is

dollars per year per unit. Given that the fixed storage cost is $1000 per year, what is the maximum possible daily profit?

The same supermarket determines that the marginal cost of storing \(q\) bouquets of flowers is approximately

dollars per month, with a fixed cost of $500 per month. Determine the total monthly cost of storing 500 bouquets a week.

A microbiologist is in charge of counting bacteria in a Petri dish. Starting with a single bacterium, she determines that the organisms are increasing at a rate of

for \(t\in [0,5]\) hours.

1. Determine as a function of \(t\) the number of organisms in the Petri dish.

   Answer
   \(N(t)=\frac{-2}{3}t^3+5t^2+100t+1\) for \(0\leq t\leq 5\text{.}\)
   Solution

   We first integrate \(N'(t)\text{:}\)

   \begin{equation*} \begin{split} N(t) \amp = \int N'(t)\,dt = \int \left(-2t^2 + 10t + 100\right)\,dt\\ \amp = -\frac{2}{3}t^3 + 5t^2 + 100t + C, \text{ for } t\in[0,5]. \end{split} \end{equation*}

   Since \(N(0)=1\text{,}\) we must have \(C=1\text{.}\) Thus, the number of organisms in the Petri dish at any time in the 5 hour period is given by

   \begin{equation*} N(t)=-\frac{2}{3}t^3 + 5t^2 + 100t + 1\text{.} \end{equation*}

2. Over the entire interval studied, what was the average number of organisms in the dish?

   Answer
   272
   Solution

   The total number of organisms over the 5 hour period is approximated by

   \begin{equation*} \int_0^5 N(t)\,dt = \int_0^5 \left(-\frac{2}{3}t^3 + 5t^2 + 100t + 1\right)\,dt = \frac{8155}{6}\text{.} \end{equation*}

   Therefore, on average, there were

   \begin{equation*} \frac{1}{5}\int_0^5 N(t)\,dt = \frac{1}{5} \frac{8155}{6} \approx 272 \text{ organisms }\text{.} \end{equation*}