Homework 8 - Solution

Date: MATH 1100-2 - Spring 2002

1. 12.1.1. If $f'(x)=4x^3$, in order to find $f$, we have to increase the power of $x$ and also divide by the increased power, that is, a possibility is $f(x)=x^4$. In fact, all the possible solutions are $f(x)=x^4+C$ for some constant $C$.

2. 12.1.3. If $f'(x)=x^6$, in order to find $f$, we have to increase the power of $x$ and also divide by the increased power, that is, a possibility is $f(x)=\frac{x^7}{7}$. In fact, all the possible solutions are $f(x)=\frac{x^7}{7}+C$ for some constant $C$.

3. 12.1.5. Using the rule $\int x^n dx = \frac{x^{n+1}}{n+1} +C$ we have: $\int x^7 dx = \frac{x^8}{8} +C$.

   Verification: $(\frac{x^8}{8} +C)' = (\frac{x^8}{8})'+0 = \frac{1}{8}\cdot 8 \cdot x^7 = x^7$.

4. 12.1.11. We first split and then use the rule $\int x^n dx = \frac{x^{n+1}}{n+1} +C$:

   $$\int(3-x^{3/2}) dx = \int 3dx - \int x^{3/2} dx = 3\int dx - \frac{x^{5/2}}{5/2} = 3x - \frac{2x^{5/2}}{5} +C.$$

   
   

   Verification: $(3x - \frac{2x^{5/2}}{5} +C)' = (3x)' - (\frac{2x^{5/2}}{5})' +0 = 3 -\frac{2}{5}\cdot \frac{5}{2} x^{3/2}$.

5. 12.1.15. We first split and then use the rule $\int x^n dx = \frac{x^{n+1}}{n+1} +C$:

   $$\begin{split}\int(2+2\sqrt{x}) dx &= \int 2dx + \int 2\sqrt{x... ...^{3/2}}{3/2} +C \\ &= 2x + \frac{4}{3} x^{3/2} +C. \end{split}$$

   
   

   Verification: $(2x + \frac{4}{3} x^{3/2} +C)' = (2x)' + (\frac{4}{3} x^{3/2})' + 0 = 2 +\frac{4}{3} \cdot \frac{3}{2} x^{1/2} = 2 + 2 x^{1/2}$.

6. 12.1.27.

   $$\begin{split}\int(x+5)^2 x dx &=\int(x^2+10x+25)x dx = \int(x... ...frac{x^4}{4} +\frac{10x^3}{3} + \frac{25x^2}{2} +C. \end{split}$$

   
   

7. 12.1.43. Since $MR(x) = R'(x)$, we have $R(x) = \int MR(x) dx = \int (3x+1) dx = 3\int xdx + \int dx = \frac{3}{2}x^2 + x +C$. But, we know that $R(0) = 0$, so that $0 = R(0) = 0+0+C$ and $C=0$. Therefore $R(x) = \frac{3}{2}x^2 + x$.

   The revenue produced by $x=50$ units is computed by $R(50) = \frac{3}{2}(50)^2 + 50 = 3800$.

8. 12.2.1. Using the substitution $u=x^2+3$, so that $du = 2xdx$ we have

   $$\int(x^2+3)^3 2x dx = \int u^3 du = \frac{u^4}{4} +C= \frac{(x^2+3)^4}{4} +C.$$

   
   

   Verification:

   $$\frac{d}{dx} (\frac{(x^2+3)^4}{4} +C) = \frac{d}{dx} (\underbrace... ...ac{d\frac{u^4}{4}}{du}\cdot \frac{d(x^2+3)}{dx} = u^3 \cdot 2x = ( x^2+3)^3 2x.$$

   
   

9. 12.2.7. We use the substitution $u=x^2+5$ so that $du = 2xdx$. We see that we have to multiply by $\frac{2}{2}$:

   $$\int(x^2+5)^3 x dx = \int (x^2+5)^3 \frac{2}{2} x dx = \int \frac... ...\int u^3 du = \frac{1}{2}\cdot\frac{u^4}{4} + C = \frac{1}{8}\cdot(x^2+5)^4 +C.$$

   
   

   Verification:

   $$\frac{d}{dx} (\frac{1}{8}\cdot(x^2+5)^4 +C) = \frac{1}{8}\frac{d}... ...{dv} \cdot \frac{d (x^2+5)}{dx} = \frac{1}{8}\cdot 4 v^3 \cdot 2x = (x^2+5)^3x.$$

   
   

10. 12.2.9. We use the substitution $u=4x-1$ so that $du = 4dx$. We see that we have to multiply by $\frac{4}{4}$:

    $$\int 7(4x-1)^6 dx = 7\int (4x-1)^6 \frac{4}{4}dx = \frac{7}{4} \int u^6 du = \frac{7}{4}\cdot \frac{u^7}{7} +C = \frac{(4x-1)^7}{4} +C.$$

    
    

    Verification:

    $$\frac{d}{dx} (\frac{(4x-1)^7}{4} +C) = \frac{1}{4}\frac{d}{dx} (\... ... \frac{dv^7}{dv}\cdot \frac{d(4x-1)}{dx} = \frac{1}{4}7 v^6\cdot4 = 7 (4x-1)^6.$$

    
    

11. 12.2.17. We use the substitution $u=x^4+6$ so that $du = 4x^3dx$. We see that we have to multiply by $\frac{4}{4}$:

    $$\int 7x^3\sqrt{x^4+6}dx = 7 \int x^3(x^4+6)^{1/2} \frac{4}{4} dx ... ...2} du = \frac{7}{4}\cdot \frac{u^{3/2}}{3/2} +C = \frac{7}{6} (x^4+6)^{3/2} +C.$$

    
    

    Verification:

    $$\begin{split}\frac{d}{dx} (\frac{7}{6} (x^4+6)^{3/2} +C) &= \... ...\cdot v^{1/2}\cdot 4x^3 \\ &= 7 (x^4+6)^{1/2} x^3. \end{split}$$

    
    

12. 12.2.27. We use the substitution $u=2x^5-5$ so that $du = 10x^4dx$. We see that we have to multiply by $\frac{10}{10}$:

    $$\begin{split}\int \frac{3x^4dx}{(2x^5-5)^4} &= 3\int \frac{10... ...5-5)^{-3}}{-3} +C = -\frac{1}{10}(2x^5-5)^{-3} +C . \end{split}$$

    
    

    Verification:

    $$\begin{split}\frac{d}{dx} (-\frac{1}{10}(2x^5-5)^{-3} +C) &= ... ...cdot (-3) v^{-4} \cdot 10 x^4 = 3(2x^5-5)^{-4} x^4. \end{split}$$

    
    

13. 12.2.39. Given the marginal revenue $MR(x) = \frac{-30}{(2x+1)^2} + 30$, we find the total revenue by integration. We will use the substitution $u=2x+1$, with $du = 2dx$:

    $$\begin{split}R(x) &= \int MR(x) dx = \int (\frac{-30}{(2x+1)^... ...frac{u^{-1}}{-1} +30 x +C = 15 (2x+1)^{-1} +30x +C. \end{split}$$

    
    
    To determine $C$ we impose the condition $R(0) = 0$, so that $0 = R(0) = 15 (2\cdot0+1)^{-1} +30\cdot 0 +C = 15\cdot 1 +C = 15+C$, so that $C=-15$. Thus, the total revenue function is $R(x) = 15 (2x+1)^{-1} +30x -15$.

14. 12.3.1. We use the substitution $u=x^3+4$, which leads to $du = 3x^2 dx$:

    $$\int\frac{3x^2}{x^3+4} dx = \int \frac{1}{u} du = \ln(\vert u\vert)+C = \ln(\vert x^3+4\vert) + C.$$

    
    

15. 12.3.3. We use the substitution $u=4z+1$, which leads to $du = 4 dz$. Along the way, we multiply the integrand by $\frac{4}{4}$:

    $$\int\frac{dz}{4z+1} = \int\frac{4}{4}\cdot\frac{dz}{4z+1} = \frac... ...u}{u} = \frac{1}{4} \ln(\vert u\vert) +C = \frac{1}{4} \ln(\vert 4z+1\vert) +C.$$

    
    

16. 12.3.17. We use the substitution $u=3x$, which leads to $du = 3dx$:

    $$\int 3 e^{3x} dx = \int e^u du = e^u +C = e^{3x}+C.$$

    
    

17. 12.3.19. We use the substitution $u=-x$, which leads to $du = -dx$. Along the way, we multiply the integrand by $\frac{-1}{-1}$:

    $$\int e^{-x} dx = \int e^{-x} \cdot\frac{-1}{-1} dx = -\int e^u du = -e^u +C = -e^{-x} +C.$$

    
    

18. 12.3.31. We first split and then use the substitutions $u=4x$ and $v=-x/2$, which lead to $du = 4dx$ and $dv = -\frac{1}{2}dx$:

    $$\begin{split}\int(e^{4x}-\frac{3}{e^{x/2}})dx &= \int e^{4x} ... ...^u + 6 e^v +C = \frac{1}{4} e^{4x} + 6 e^{-x/2} +C. \end{split}$$

    
    

19. 12.4.1. The total cost is the integral of the marginal cost:

    $$\begin{split}C(x) &= \int MC(x) dx = \int (2x+100) dx = \int ... ... = 2\int x dx + 100 \int dx \\ &= x^2 + 100 x + K. \end{split}$$

    
    
    To determine the constant $K$ we use that the fixed costs are $100$, that is, $100 = C(0) = 0+0+K$, so that $K=100$ and the total cost function is $C(x) = x^2 + 100 x +100$.

20. 12.4.7. We start by finding the total revenue $R(x)$:

    $$\begin{split}R(x) &= \int MR(x) dx = \int (44-5x) dx = \int 4... ...nt dx - 5\int xdx \\ &= 44 x - \frac{5}{2} x^2 + K \end{split}$$

    
    
    To find $K$ we use that $R(0) = 0$, so that $0 = R(0) = 0 - 0 +K$ and $K=0$. Therefore $R(x)= 44 x - \frac{5}{2} x^2$.

    Next we find the total cost:

    $$\begin{split}C(x) &= \int MC(x) dx = \int (3x+20)dx = \int 3x... ...int x dx + 20\int dx \\ &= \frac{3}{2}x^2 +20x +K. \end{split}$$

    
    
    To find $K$ we use that $R(80) = 11400$, so that $11400 = R(80) = \frac{3}{2} 80^2 +20\cdot80 +K$, so that $11400 = 11200 + K$, and $K=200$. Therefore, $C(x) = \frac{3}{2}x^2 +20x+200$.

    The profit function is $P(x) = R(x)-C(x) = 44 x - \frac{5}{2} x^2 - (\frac{3}{2}x^2 +20x+200) = -4x^2 +24x -200$.

    The maximum profit is achieved when $P'(x)=0$, that is, when $-8x+24=0$, or $x=3$ (actually, since $P''(3)=-8<0$, we know that $x=3$ is a maximum).

    Finally, the maximum profit is $P(3) = -164<0$, so that there is a loss at the optimum level.