Homework 9 - Solution

Date: MATH 1100-2 - Spring 2002

1. 12.5.3. Since $y=3x^2+1$, $dy = y' dx = 6x dx$. Then

   $$2ydx-xdy = 2 (3x^2+1) dx - x \cdot 6x dx = 6x^2dx + 2dx - 6x^2dx = 2dx,$$

   
   
   as wanted.

2. 12.5.7. To solve $2ydy=4xdx$, since the variables are separated we have to integrate both sides:

   $$2ydy=4xdx \ensuremath{\Rightarrow}\xspace \int 2ydy= \int 4xdx \ensuremath{\Rightarrow}\xspace y^2 = 4\frac{x^2}{2} +C = 2x^2+C.$$

   
   
   Then, the general solution (in implicit form) is $y^2=2x^2+C$. The explicit form is $y=\pm\sqrt{2x^2+C}$.

3. 12.5.15. For $\frac{dy}{dx} = \frac{x^2}{y}$, we start by separating the variables: $ydy = x^2dx$ and then integrate both sides:

   $$ydy = x^2dx \ensuremath{\Rightarrow}\xspace \int ydy = \int x^2dx \ensuremath{\Rightarrow}\xspace \frac{y^2}{2} = \frac{x^3}{3} + C,$$

   
   
   so that the general solution (in implicit form) is $\frac{y^2}{2} = \frac{x^3}{3} + C$. The explicit form of the solution is: $y = \pm\sqrt{\frac{2}{3}x^3+C'}$.

4. 12.5.29. For $\frac{dy}{dx} = \frac{x^2}{y^3}$, we start by separating the variables: $y^3dy = x^2dx$ and then integrate both sides:

   $$y^3dy = x^2dx \ensuremath{\Rightarrow}\xspace \int y^3dy = \int x^2dx \ensuremath{\Rightarrow}\xspace \frac{y^4}{4} = \frac{x^3}{3} + C,$$

   
   
   so that the general solution (in implicit form) is $\frac{y^4}{4} = \frac{x^3}{3} + C$. To find the particular solution that has $y=1$ when $x=1$ we evaluate at these points and solve for $C$:

   $$\frac{1^4}{4} = \frac{1^3}{3} + C \ensuremath{\Rightarrow}\xspace C = \frac{1}{4} - \frac{1}{3} = -\frac{1}{12}.$$

   
   
   All together, the particular solution is (in implicit form) $\frac{y^4}{4} = \frac{x^3}{3} - \frac{1}{12}$.

5. 13.1.16.

   $$\sum_{i=3}^5 (i^2+1) = (3^2+1) + (4^2+1) + (5^2+1) = 10 + 17 + 26 = 53.$$

   
   

6. 13.1.21.

   $$\sum_{j=1}^{60} 3 = 3\sum_{j=1}^{60} 1 = 3\cdot 60 = 180.$$

   
   

7. 13.1.23.

   $$\begin{split}\sum_{k=1}^{30} (k^2+4k) &= \sum_{k=1}^{30} k^2 ... ... + 4 \frac{30\cdot (30+1)}{2} = 9455 + 1860 = 11315 \end{split}$$

   
   

8. 13.2.1. We have $\int 4x dx = 4\int x dx = 4 \frac{x^2}{2} = 2x^2$ (notice that we don't write the constant $C$ because it is not needed for the computation of the definite integral):

   $$\int_0^3 4x dx = 2x^2 \vert _0^3 = 2\cdot 3^2 - 2\cdot 0^2 = 2\cdot 9 - 0= 18.$$

   
   

9. 13.2.7. Since $\int 4\sqrt[3]{x^2} dx = 4\int x^{2/3} dx = 4 \cdot \frac{3}{5}\cdot x^{5/3} = \frac{12}{5} x^{5/3}$, we have:

   $$\int_0^5 4\sqrt[3]{x^2} dx = \frac{12}{5} x^{5/3}\vert _0^5 = \fr... ...^{5/3} - \frac{12}{5} \cdot 0^{5/3} =\frac{12}{5} \cdot 5^{5/3} \simeq 35.0882.$$

   
   

10. 13.2.13. Using the substitution $u=x^2+2$, so that $du = 2xdx$ we have $\int (x^2+2)^3 x dx = \frac{1}{2} \int u^3 du = \frac{u^4}{8} = \frac{1}{8} (x^2+2)^4$. Then:

    $$\int_2^4 (x^2+2)^3 x dx = \frac{1}{8} (x^2+2)^4\vert _2^4 = \frac{1}{8} (4^2+2)^4 - \frac{1}{8} (2^2+2)^4 = 13122 - 162 = 12960.$$

    
    

11. 13.2.23. Since $\int \frac{1}{z} dz = \ln(\vert z\vert)$, we have

    $$\int_1^e \frac{1}{z}dz = \ln(\vert z\vert)\vert _1^e = \ln(\vert e\vert) - \ln(\vert 1\vert) = 1 - 0 = 1.$$

    
    

12. 13.2.34.
    1. We want the area delimited by $f(x) = \frac{1}{2} x^2 +x+1$, the $x$-axis, $x=-2$ and $x=1$. This area is computed by the definite integral $\int_{-2}^1 (\frac{1}{2} x^2 +x+1) dx$.
    2. We first compute $\int (\frac{1}{2} x^2 +x+1) dx = \frac{1}{2} \int x^2 dx + \int xdx + \int dx = \frac{x^3}{6} + \frac{x^2}{2} +x$. Then:

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