MATH 16B SECOND MIDTERM SOLUTIONS

Prepared by Di-an Jan

1. 1. First, try to simplify the integrand, in this case, with one of the laws of exponents

      $$\int_2^3 x^2 (e^x)^2 \,d x= \int_2^3 x^2 e^{2x} \,d x$$

      
      
      this is a product but not in the form of a substitution, so try integration by parts. Now the derivative of $x^2$ is simpler and the antiderivative is more complicated, while the derivative and antiderivatives of $e^{2x}$ are about the same in complexity, so try $$\begin{array}[t]{r@{}l@{\quad}r@{}l} f (x) &= x^2 & g(x) &= ... ...\\ f'(x) &= 2x & G(x) &= \tfrac{1}{2}e^{2x} \\ \end{array}$$

      $$\int x^2 e^{2x} \,d x= \tfrac{1}{2} x^2 e^{2x} - \int x e^{2x} \,d x$$

      
      
      To do the resulting integral, again integrate by parts $$\begin{array}[t]{r@{}l@{\quad}r@{}l} f (x) &= x & g(x) &= e^... ... \\ f'(x) &= 1 & G(x) &= \tfrac{1}{2}e^{2x} \\ \end{array}$$

      $$\int x e^{2x} \,d x = \tfrac{1}{2} x e^{2x} - \tfrac{1}{2} \int e^{2x} \,d x = \tfrac{1}{2} x e^{2x} - \tfrac{1}{4} e^{2x} + C$$

      
      
      Finally, we put it all together with the limits,

      $$\int_2^3 x^2 (e^x)^2 \,d x = \Bigl(\tfrac{1}{2} x^2 e^{2x} - \bigl(\tfrac{1}{2} x e^{2x} - \tfrac{1}{4} e^{2x}\bigr) \Bigr\vert _2^3$$

      $$= \Bigl(\tfrac{1}{2} x^2 e^{2x} - \tfrac{1}{2} x e^{2x} + \tfrac{1}{4} e^{2x} \Bigr\vert _2^3 = \tfrac{13}{4} e^6 - \tfrac{5}{4} e^4$$

      
      

   2. Again, first simplify the integrand

      $$\int_0^1 \dfrac{\ln (\sqrt[4]{x})}{x} \,d x = \int_0^1 \dfrac{\ln (x^{1/4})}{x} \,d x = \int_0^1 \dfrac{\ln x}{4x} \,d x$$

      
      
      Note that this is an improper integral because both $\ln x$ and  $\tfrac{1}{x}$ are not defined at $x = 0$. To do the integral, note that the derivative of $\ln x$ is  $\tfrac{1}{x}$, so do substitution with $u = \ln x$ and  $du = \tfrac{1}{x} \,d x$

      $$\int \dfrac{\ln (\sqrt[4]{x})}{x} \,d x = \tfrac{1}{4} \int u \,d u = \tfrac{1}{8} u^2 + C = \tfrac{1}{8} (\ln x)^2 + C$$

      
      
      But we need to plug in the lower limit as an improper integral

      $$\int_0^1 \dfrac{\ln (\sqrt[4]{x})}{x} \,d x = \lim_{a \to 0^+} \int_a^1 \dfrac{\ln (\sqrt[4]{x})}{x} \,d x$$

      $$= \lim_{a \to 0^+} \tfrac{1}{8} \bigl((\ln 1)^2 - (\ln a)^2\bigr)$$

      $$= \lim_{a \to 0^+} - \tfrac{1}{8} (\ln a)^2$$

      
      
      Note that $\ln 1 = 0$. To evaluate the limit, note that as $x \to 0$ from the positive side, $\ln x$ becomes very big negative, so $(\ln x)^2$ becomes very big positive, and  $-\tfrac{1}{8} (\ln a)^2$ becomes very big negative. Thus the integral diverges, and we say

      $$\int_0^1 \dfrac{\ln(\sqrt[4]{x})}{x} \,d x= -\infty$$

      
      

      There are other ways to do the integral correctly. These all give the same answer after simplifying with laws of logarithms and exponents.

      1. Substitution with  $u = \ln (\sqrt[4]{x}) = \ln \bigl(x^{1/4}\bigr)$ and  $du = \tfrac{1}{x^{1/4}} \bigl(\tfrac{1}{4} x^{-3/4}\bigr) \,d x = \tfrac{1}{4x} \,d x$

         $$\int \dfrac{\ln(\sqrt[4]{x})}{x} \,d x = 4 \int u \,d u= 2 u^2 + C = 2 \bigl(\ln(\sqrt[4]{x})\bigr)^2 + C$$

         
         
         Do not forget the chain rule in computing $du$.

      2. Integration by parts with $$\begin{array}[t]{r@{}l@{\quad}r@{}l} f (x) &= \ln(\sqrt[4]{x... ... \\ f'(x) &= \tfrac{1}{4x} & g(x) &= \ln(x) \\ \end{array}$$

         $$\int \dfrac{\ln(\sqrt[4]{x})}{x} \,d x = \ln(\sqrt[4]{x}) \ln(x) - \int \dfrac{\ln x}{4x} \,d x$$

         $$= \ln(\sqrt[4]{x}) \ln(x) - \tfrac{1}{8}(\ln x)^2 + C$$

         
         
         where $f'(x)$ and  $\int \tfrac{\ln x}{4x} \,d x$ are as calculated above.

         Integrating by parts don't work the other way because we don't know the antiderivative of  $\ln(\sqrt[4]{x})$.

      3. Same as above but integrate  $\int \tfrac{\ln x}{x} \,d x$ by parts $$\begin{array}[t]{r@{}l@{\quad}r@{}l} f (x) &= \ln x & G(x) &... ...x} \\ f'(x) &= \tfrac{1}{x} & g(x) &= \ln x \\ \end{array}$$

         $$\int \dfrac{\ln x}{x} \,d x= (\ln x)^2 - \int \dfrac{\ln x}{x} \,d x$$

         
         
         Note that the two integrals are the some, so moving them to the same side gives

         $$\int \dfrac{\ln x}{x} \,d x= \tfrac{1}{2} (\ln x)^2$$

         
         

         Integrating by parts the other way don't work. It gives

         $$\int \dfrac{\ln x}{x} \,d x = \ln x - 1 + \int \dfrac{\ln x}{x} \,d x- \int \dfrac{1}{x} \,d x = C - 1 + \int \dfrac{\ln x}{x} \,d x$$

         
         
         because  $\int \tfrac{1}{x} \,d x= \ln x + C$. This don't tell us anything as indefinite integrals differ by a constant.

   3. There are a couple ways to do $\int \cos(x) \sin(x) \sqrt[3]{\sin(x) + 2} \,d x$

      1. The integrand contains a single  $\cos x \,d x$ and many $\sin x$, whose derivative is $\cos x$, so try the substitution  $u = \sin x$ and  $du = \cos x \,d x$

         $$\int \cos(x) \sin(x) \sqrt[3]{\sin(x) + 2} \,d x = \int u (u + 2)^{1/3} \,d u$$

         
         
         then all we can do next is integration by parts. Since the derivative of $u$ is the simplest, we should try $$\begin{array}[t]{r@{}l@{\quad}r@{}l} f (u) &= u & g(u) &= (u... ...'(u) &= 1 & G(u) &= \tfrac{3}{4} (u + 2)^{4/3} \\ \end{array}$$

         $$\int u (u+2)^{1/3} \,d u = \tfrac{3}{4} u (u + 2)^{4/3} - \tfrac{3}{4} \int (u + 2)^{4/3} \,d u$$

         $$= \tfrac{3}{4} u (u + 2)^{4/3} - \tfrac{9}{28} (u + 2)^{7/3} + C$$

         $$= \tfrac{3}{4} \sin(x) \, \bigl(\sin(x) + 2\bigr)^{4/3} - \tfrac{9}{28} \bigl(\sin(x) + 2\bigl)^{7/3} + C$$

         
         
         Note that the variable is $u$, which is converted back to $x$ at the end.

      2. Instead, we can take a bigger substitution without changing its derivative  $u = \sin(x) + 2$ and  $du = \cos x \,d x$

         $$\int \cos(x) \sin(x) \sqrt[3]{\sin(x) + 2} \,d x = \int (u - 2) u^{1/3} \,d u$$

         
         
         Note that we used  $u = \sin(x) + 2$ to replace  $\sin(x) = u - 2$. This is necessary because $x$ depends on $u$ because $u$ depends on $x$, and when we integrate with respect to $u$ we need to know exactly how $\sin(x)$ changes with respect to $u$. There are two ways to do the integral:

         1. Integrate by parts, with $f(x)$ chosen to have the simplest derivative $$\begin{array}[t]{r@{}l@{\quad}r@{}l} f (u) &= u - 2 & g(u) &... ...\ f'(u) &= 1 & G(u) &= \tfrac{3}{4} u^{4/3} \\ \end{array}$$

            $$\int (u - 2) u^{1/3} \,d u = \tfrac{3}{4}(u - 2) u^{4/3} - \tfrac{3}{4} \int u^{4/3} \,d u$$

            $$= \tfrac{3}{4}(u - 2) u^{4/3} - \tfrac{9}{28} u^{7/3} + C$$

            
            
            which is the same as before when we put in  $u = \sin x + 2$.

         2. Or, we can multiply it out and use the laws of exponents to simplify

            $$\int (u - 2) u^{1/3} \,d u = \int \bigl(u^{4/3} - 2 u^{1/3}\bigr) \,d u$$

            $$= \tfrac{3}{7} u^{7/3} - \tfrac{3}{2} u^{4/3} + C$$

            $$= \tfrac{3}{7} \bigl(\sin(x) + 2\bigr)^{7/3} - \tfrac{3}{2} \bigl(\sin(x) + 2\bigr)^{4/3} + C$$

            
            
            and we can see this is the same by multiplying out $\tfrac{3}{4}(u - 2) u^{4/3} - \tfrac{9}{28} u^{7/3}$

      3. We can also try integration by parts first, but it's harder to see how to choose $f(x)$ and $g(x)$ without before using substitution to simplify the integral. For $g(x)$, we need something we can integrate to $G(x)$

         $$\begin{array}{r@{}l@{\quad}r@{}l} g(x) &= \sin(x) & G(x) &= -... ...) &= \tfrac{3}{4} \bigl(\sin(x) + 2\bigr)^{4/3} \\ \end{array}$$

         
         
         The last case corresponds to  $f(x) = \sin(x)$ and  $f'(x) = \cos(x)$. It is the only one where $f'(x)$ is no more complicated than $f(x)$ and in fact, the only one that can be done by integrating by parts, giving

         $$\int \cos(x) \sin(x) \sqrt[3]{\sin(x) + 2} \,d x = \tfrac{3}{4} \sin(x) \, \bigl(\sin(x) + 2\bigr)^{4/3} - \tfrac{3}{4} \int \cos(x) \, \bigl(\sin(x) + 2\bigr)^{4/3} \,d x$$

         $$= \tfrac{3}{4} \sin(x) \, \bigl(\sin(x) + 2\bigr)^{4/3} - \tfrac{9}{28} \bigl(\sin(x) + 2\bigr)^{7/3} + C$$

         
         

2. To approximate  $\int_0^4 2^x \,d x$ using Simpson's rule with $n = 2$, divide the interval $[0,4]$ into $2$ pieces bounded by the points $a_0 = 0$, $a_1 = 2$ and $a_2 = 4$ each of size  $\Delta x= 2$. Let  $f(x) = 2^x$, then the trapezoid rule approximation is

   $$T_2 = \left(\frac{f(a_0) + f(a_1)}{2} + \frac{f(a_1) + f(a_2)}{2}\right) \Delta x$$

   $$= \frac{f(a_0) + 2 f(a_1) + f(a_2)}{2} \Delta x$$

   $$= 1 + 8 + 16 = 25$$

   
   
   The midpoints are $x_1 = 1$ and $x_2 = 3$ so the midpoint rule approximation is

   $$M_2 = \bigl(f(x_1) + f(x_2)\bigr) \Delta x$$

   $$= (2 + 8)(2) = 20$$

   
   
   The Simpson rule approximation is

   $$S_2 = \frac{2 M_2 + T_2}{3} = \frac{65}{3} = 21\tfrac{2}{3}$$

   
   
   Or, with the other formula,

   $$S_2 = \dfrac{f(a_0) + 4 f(x_1) + 2 f(a_1) + 4 f(x_2) + f(a_2)}{6} \Delta x$$

   $$= \dfrac{1 + 8 + 8 + 32 + 16}{3} = \frac{65}{3} = 21\tfrac{2}{3}$$

   
   

3. To solve the differential equation  $y' = t e^{t^2} y$, with initial condition  $y(0) = -10$, use separation of variables. First, there is a constant solution $y = 0$, but that doesn't satisfy the initial condition. Then

   $$\int \dfrac{y'}{y} \,d t = \int t e^{t^2} \,d t$$

   $$\int \dfrac{1}{y} \,d y = \tfrac{1}{2} \int e^u \,d u$$

   $$\ln\lvert y \rvert = \tfrac{1}{2} e^{t^2} + C$$

   $$\lvert y \rvert = e^{\frac{1}{2} e^{t^2} + C} = e^{\frac{1}{2} e^{t^2}} e^C = A e^{\frac{1}{2} e^{t^2}}$$

   
   
   where $u = t^2$ and $A = e^C$. Since  $y(0) = -10$ is negative, we must have $y(t) < 0$ and  $\lvert y \rvert = -y$

   $$y = -A e^{\frac{1}{2} e^{t^2}}$$

   
   
   The initial condition requires

   $$-10 = y(0) = -A e^{\frac{1}{2} e^{0^2}} = -A e^{\frac{1}{2}} = -A \sqrt{e}$$

   
   
   so  $A = \tfrac{10}{\sqrt{e}}$ and

   $$y(t) = -\tfrac{10}{\sqrt{e}} e^{\frac{1}{2} e^{t^2}} = -10 e^{\frac{1}{2} (e^{t^2}-1)}$$

   
   

4. A Euler's method problem should be on the final.

5. 1. Start at $t = 0$ at the given initial value $y(0) = -1$. There, the slope  $y' = g(y) = g(-1)$ is positive, so $y$ increases. Thus we move right (the positive direction) along the $y$-axis, and we find that the slope $y' = g(y)$ decreases until it reaches 0 at $y = 0$. Thus the solution $y(t)$ starts from $t = 0$ at $y = -1$ with a positive slope, and concaves up and becomes asymptotic to $y = 0$. This is (d).

   2. Start at $t = 0$ at the given initial value $y(0) = 1$. There, the slope  $y' = g(y) = g(1)$ is negative, so $y$ decreases. Thus we move left (the negative direction) along the $y$-axis, and we find that the slope $y' = g(y)$ increases (gets closer to 0) until it reaches 0 at $y = 0$. Thus the solution $y(t)$ starts from $t = 0$ at $y = 1$ with a negative slope, and concaves down and becomes asymptotic to $y = 0$. This is (b).

6. Let $y$ be the amount of compound A, then the rate at which A transforms to B is $y'$. The rate is negative because the amount of A decreases as it is transformed to B, but that's okay. ``The rate is proportional to the cube of B,'' and the other conditions translates to

   $$y' = k y^3 \qquad y(0) = 20 \qquad y(1) = 15 \\$$

   
   
   for some (negative) constant $k$. The differential equation can be easily solved by separation of variables. There is a constant solution $y = 0$ but that doesn't work, so

   $$\int y^{-3} y' \,d t = \int k \,d t$$

   $$\int y^{-3} \,d y = \int k \,d t$$

   $$-\tfrac{1}{2} y^{-2} = k t + C$$

   $$y^{-2} = -2 k t - 2 C$$

   $$y = (-2 k t - 2 C)^{-1/2} = \frac{1}{\sqrt{-2 k t - 2 C}}$$

   
   
   Initially, there are $20$ moles of A, so plug in $t = 0$ and $y = 20$, and the next to last equation above gives

   $$20^{-2} = 0 - 2 C \qquad -2 C = \tfrac{1}{400}$$

   
   

   $$y = \frac{1}{\sqrt{\tfrac{1}{400} - 2 k t}} = \frac{20}{\sqrt{1 - 800 k t}}$$

   
   
   After $1$ second, there are $15$ moles of A, so plug in $t = 1$ and $y = 15$

   $$15 = \frac{20}{\sqrt{1 - 800 k}}$$

   $$\tfrac{3}{4} = \frac{1}{\sqrt{1 - 800 k}}$$

   $$1 - 800 k = \tfrac{16}{9}$$

   $$-800 k = \tfrac{7}{9}$$

   
   
   The solution is

   $$y = \frac{20}{\sqrt{1 + \tfrac{7}{9} t}} = \frac{60}{\sqrt{9 + 7 t}}$$

   
   
   The amount of A after $2$ seconds is

   $$y(2) = \frac{60}{\sqrt{23}}$$