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Chapter 5: Problem 22

Consider the following functions \(f\) and real numbers a (see figure). a. Find and graph the area function \(A(x)=\int_{a}^{x} f(t) d t\) b. Verify that \(A^{\prime}(x)=f(x)\) $$f(t)=4 t+2, a=0$$

Short Answer

Expert verified
Question: Determine the area function A(x) for the given function f(t) = 4t + 2 and starting point a = 0, and verify that A'(x) = f(x). Answer: The area function A(x) for the given function f(t) = 4t + 2 and starting point a = 0 is A(x) = 2x^2 + 2x. It has been verified that A'(x) = f(x) = 4x + 2.

Step by step solution

01

Find the area function A(x)

To find the area function \(A(x)\), we will integrate the given function \(f(t)=4t+2\) with respect to t, from a to x: $$A(x) = \int_{a}^{x} f(t) dt = \int_{0}^{x} (4t + 2) dt$$
02

Integrate the function f(t)

Now, we will integrate the function \((4t + 2)\) with respect to t: $$\int (4t + 2) dt = 4\int t dt + 2\int dt$$ $$= 4(\frac{1}{2}t^2) + 2t + C$$ $$= 2t^2 + 2t + C$$ However, since the area function starts at 0, we can determine the constant C: $$A(0) = 2(0)^2 + 2(0) + C = 0$$ $$C = 0$$ So the area function is: $$A(x) = 2x^2 + 2x$$
03

Graph the area function A(x)

To graph the area function, plot the function \(A(x) = 2x^2 + 2x\) on a graphing tool. The graph represents the accumulated area under the curve of the function \(f(t)=4t+2\) from a=0 to any given point x.
04

Verify A'(x)=f(x)

Find the derivative of the area function A(x) with respect to x: $$A'(x) = \frac{d(2x^2 + 2x)}{dx}$$ $$= 4x + 2$$ We can see from the result that \(A'(x) = f(x)\). Therefore, it is verified that \(A'(x) = f(x) = 4x + 2\).

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Most popular questions from this chapter

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