4.10 Antiderivatives  (Page 4/10)

1. Using [link] , we can integrate each of the four terms in the integrand separately. We obtain
   
   ${\displaystyle \int \left(5x^3-7x^2+3x+4\right)dx}={\displaystyle \int 5x^{3}dx-}{\displaystyle \int 7x^{2}dx}+{\displaystyle \int 3xdx}+{\displaystyle \int 4dx}.$
   
   From the second part of [link] , each coefficient can be written in front of the integral sign, which gives
   
   ${\displaystyle \int 5x^{3}dx-}{\displaystyle \int 7x^{2}dx}+{\displaystyle \int 3xdx}+{\displaystyle \int 4dx}=5{\displaystyle \int x^{3}dx-7}{\displaystyle \int x^{2}dx}+3{\displaystyle \int xdx}+4{\displaystyle \int 1dx.}$
   Using the power rule for integrals, we conclude that
   
   ${\displaystyle \int \left(5x^3-7x^2+3x+4\right)dx}=\frac{5}{4}{x}^4-\frac{7}{3}{x}^3+\frac{3}{2}{x}^2+4x+C.$
2. Rewrite the integrand as
   
   $\frac{x^2+4\sqrt[3]{x}}{x}=\frac{x^2}{x}+\frac{4\sqrt[3]{x}}{x}=0.$
   
   Then, to evaluate the integral, integrate each of these terms separately. Using the power rule, we have
   
   $\begin{array}{cc}\hfill {\displaystyle \int \left(x+\frac{4}{x^{2\text{/}3}}\right)}dx& ={\displaystyle \int xdx+4}{\displaystyle \int x^{\mathrm{-2}\text{/}3}}dx\hfill \\ & =\frac{1}{2}{x}^2+4\frac{1}{\left(\frac{\mathrm{-2}}{3}\right)+1}{x}^{\left(\mathrm{-2}\text{/}3\right)+1}+C\hfill \\ & =\frac{1}{2}{x}^2+12x^{1\text{/}3}+C.\hfill \end{array}$
3. Using [link] , write the integral as
   
   $4{\displaystyle \int \frac{1}{1+x^2}dx}.$
   
   Then, use the fact that ${\text{tan}}^{\mathrm{-1}}\left(x\right)$ is an antiderivative of $\frac{1}{\left(1+x^2\right)}$ to conclude that
   
   ${\displaystyle \int \frac{4}{1+x^2}dx}=4{\text{tan}}^{\mathrm{-1}}\left(x\right)+C.$
4. Rewrite the integrand as
   
   $\text{tan}x\text{cos}x=\frac{\text{sin}x}{\text{cos}x}\text{cos}x=\text{sin}x.$
   
   Therefore,
   
   ${\displaystyle \int \text{tan}x\text{cos}x}={\displaystyle \int \text{sin}x}=\text{−}\text{cos}x+C.$

First we need to solve the differential equation. If $\frac{dy}{dx}=\text{sin}x,$ then

Next we need to look for a solution $y$ that satisfies the initial condition. The initial condition $y\left(0\right)=5$ means we need a constant $C$ such that $\text{−}\text{cos}x+C=5.$ Therefore,

The solution of the initial-value problem is $y=\text{−}\text{cos}x+6.$