MHB Vanessa's questions at Yahoo Answers regarding applications of integration

[MarkFL]

Here are the questions:

Determine the area of the region between the curves?

(a) If 77600 dollars is deposited in an account for 10 years at 5.35 percent compounded continuously , find the average value of the account during the 10 years period.

(b) Determine the area of the region between the curves: f(x) = -1*x^2 + 10x - 6 and g(x) = 4x + 3.
--for this one I know I'm supposed to get the intersections, but I'm only getting one x-value, 3. What do I do?Thank you in advance! :)

I have posted a link there to this topic so the OP can see my work.

 

[MarkFL]

Hello Vanessa,

(a) We may model the amount $A$ in dollars in the account at time $t$ in years with the IVP:

$$\frac{dA}{dt}=rA$$ where $$A(0)=A_0$$

Separate variables, exchange dummy variables of integration and use boundaries as limits:

$$\int_{A_0}^{A(t)}\frac{du}{u}=r\int_0^t\,dv$$

Applying the FTOC, we find:

$$\left[\ln|u| \right]_{A_0}^{A(t)}=r\left[v \right]_0^t$$

$$\ln\left(\frac{A(t)}{A_0} \right)=rt$$

Convert from logarithmic to exponential form, solving for $A(t)$:

$$A(t)=A_0e^{rt}$$

Now, the average value in the account over a period of $t$ years is:

$$\overline{A(t)}=\frac{1}{t-0}\int_0^t A_0e^{ru}\,du=\frac{A_0}{rt}\int_0^t e^{ru}\,r\,du=\frac{A_0}{rt}\left[e^{ru} \right]_0^t=\frac{A_0\left(e^{rt}-1 \right)}{rt}$$

Now, using the given data:

$$A_0=77600,\,r=0.0535,\,t=10$$, we find:

$$\overline{A(10)}=\frac{77600\left(e^{0.0535\cdot10}-1 \right)}{0.0535\cdot10}\approx102613.05$$

(b) First, let's find the points of intersection:

$$f(x)=g(x)$$

$$-x^2+10x-6=4x+3$$

$$x^2-6x+9=0$$

$$(x-3)^2=0$$

You are correct, we find that $g(x)$ is tangent to $f(x)$ and so the area of region bounded by the two functions is zero.