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

AI Thread Summary
The discussion addresses two mathematical problems posed by Vanessa regarding integration applications. For the first question, the average value of an account with a $77,600 deposit at a 5.35% continuous compounding rate over 10 years is calculated to be approximately $102,613.05. In the second question, the curves f(x) = -1*x^2 + 10x - 6 and g(x) = 4x + 3 are analyzed, revealing that they intersect at a single point, indicating that g(x) is tangent to f(x), resulting in an area of zero between the curves. The calculations and reasoning are provided to clarify these concepts. The thread effectively demonstrates the application of integration in financial and geometric contexts.
MarkFL
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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.
 
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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.
 
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