Problem 1: Evaluate the following integrals.

a) $\int \frac{a{x}^2+bx+c}{x}dx$

Solution:

$$\int \frac{a{x}^2+bx+c}{x}dx=\int (ax+b+\frac{c}{x})dx=\frac{a}{2}{x}^2+bx+c\ln \mathrm{\mid }x\mathrm{\mid }+C$$

b) $\int \frac{e^{2x}+e^x+1}{e^x}dx$

Solution:

$$\int \frac{e^{2x}+e^x+1}{e^x}dx=\int (e^x+1+e^{-x})dx=e^x+x-e^{-x}+C$$

c) $\int {(\sqrt{x}+\frac{1}{\sqrt{x}})}^{2}dx$

Solution:

$$\int {(\sqrt{x}+\frac{1}{\sqrt{x}})}^{2}dx=\int (x+2+\frac{1}{x})dx=\frac{1}{2}{x}^2+2x+\ln \mathrm{\mid }x\mathrm{\mid }+C$$

Problem 2: Evaluate the following integrals.

a) ${\int }_0^{1}8x(x^2+1)dx$

Solution:

$${\int }_0^{1}8x(x^2+1)dx=8{\int }_0^1(x^3+x)dx=8{[\frac{1}{4}{x}^4+\frac{1}{2}{x}^2]}_0^1=8(\frac{1}{4}+\frac{1}{2})=8\times \frac{3}{4}=6$$

b) ${\int }_0^{1}4x^2+4dx$

Solution:

$${\int }_0^1(4x^2+4)dx=4{\int }_0^1{x}^{2}dx+4{\int }_0^{1}1dx=4{\left[\frac{1}{3}{x}^3\right]}_0^1+4{\left[x\right]}_0^1=4\times \frac{1}{3}+4\times 1=\frac{4}{3}+4=\frac{16}{3}$$

Problem 3: Differential Equations

a) $\frac{d^{2}y}{d{x}^2}=25e^{5x};f^{\mathrm{\prime }}(0)=4,f(0)=-2$

Solution:
First, integrate the second derivative to find the first derivative:

$$\frac{dy}{dx}=\int 25e^{5x}dx=5e^{5x}+C_1$$

Given $f^{\mathrm{\prime }}(0)=4$:

$$5e^0+C_1=4\Rightarrow 5+C_1=4\Rightarrow C_1=-1$$

Now, integrate to find $y$:

$$y=\int (5e^{5x}-1)dx=e^{5x}-x+C_2$$

Given $f(0)=-2$:

$$e^0-0+C_2=-2\Rightarrow 1+C_2=-2\Rightarrow C_2=-3$$

Thus, the solution is:

$$y=e^{5x}-x-3$$

Problem 4: Marginal Profit

The marginal profit for producing $x$ items is given by $MF=500-kx$. Find the total profit function if $TP=0$ when $x=0$. What is the maximum profit?

Solution:
The total profit function $TP(x)$ is the integral of the marginal profit:

$$TP(x)=\int (500-kx)dx=500x-\frac{k}{2}{x}^2+C$$

Given $TP(0)=0$:

$$0=500\times 0-\frac{k}{2}\times 0+C\Rightarrow C=0$$

Thus, the total profit function is:

$$TP(x)=500x-\frac{k}{2}{x}^2$$

To find the maximum profit, take the derivative and set it to zero:

$$\frac{dTP}{dx}=500-kx=0\Rightarrow x=\frac{500}{k}$$

Substitute $x$ back into $TP(x)$:

$$TP\left(\frac{500}{k}\right)=500\times \frac{500}{k}-\frac{k}{2}{\left(\frac{500}{k}\right)}^2=\frac{250000}{k}-\frac{125000}{k}=\frac{125000}{k}$$

Thus, the maximum profit is $\frac{125000}{k}$.

Hope this will help you.