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Chapter 5: Problem 31

What is the average distance between the parabola \(y=30 x(20-x)\) and the \(x\) -axis on the interval [0,20]\(?\)

Short Answer

Expert verified
Answer: The average distance is approximately 2000.

Step by step solution

01

Integrate the function over the given interval

In this step, we will find the integral of the function \(y=30x(20-x)\) over the interval \([0,20]\). Using the power rule, integrate the function with respect to x: \begin{align*} \int_{0}^{20} 30x(20-x) \, dx &= 30 \int_{0}^{20} (20x - x^2) \, dx \\ &= 30 \left[ \int_{0}^{20} 20x \, dx - \int_{0}^{20} x^2 \, dx \right] \\ &= 30 \left[ 10x^2 \Big|_{0}^{20} - \frac{1}{3}x^3 \Big|_{0}^{20} \right] \\ &= 30 \left[ (10 \cdot 20^2) - \frac{1}{3}(20^3) \right] \\ \end{align*}
02

Divide the integral by the length of the interval

Now we need to divide the integral found in Step 1 by the length of the interval. The interval is \([0,20]\), so the length is \(20 - 0 = 20\). We'll divide the result of the integral by 20: $$ \frac{30 \left[ (10 \cdot 20^2) - \frac{1}{3}(20^3) \right]}{20} $$
03

Simplify to find the average distance

Finally, we will simplify the expression to find the average distance between the parabola and the x-axis on the given interval: \begin{align*} \frac{30 \left[ (10 \cdot 20^2) - \frac{1}{3}(20^3) \right]}{20} &= \frac{30(10 \cdot 400 - \frac{1}{3}(8000))}{20} \\ &= \frac{30(4000 - 2666.6667)}{20} \\ &= \frac{30 \cdot 1333.3333}{20} \\ &\approx \frac{39999.999}{20} \\ &\approx 2000 \end{align*} The average distance between the parabola \(y=30x(20-x)\) and the x-axis on the interval \([0,20]\) is approximately 2000.

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Most popular questions from this chapter

The following functions describe the velocity of a car (in mi/hr) moving along a straight highway for a 3-hr interval. In each case, find the function that gives the displacement of the car over the interval \([0, t],\) where \(0 \leq t \leq 3\). $$v(t)=\left\\{\begin{array}{ll} 40 & \text { if } 0 \leq t \leq 1.5 \\ 50 & \text { if } 1.5 < t \leq 3 \end{array}\right.$$

Additional integrals Use a change of variables to evaluate the following integrals. $$\int \sec ^{2} 10 x d x$$

Use geometry to evaluate the following integrals. $$\int_{-6}^{4} \sqrt{24-2 x-x^{2}} d x$$

Use geometry and the result of Exercise 76 to evaluate the following integrals. $$\int_{1}^{6} f(x) d x, \text { where } f(x)=\left\\{\begin{array}{ll}2 x & \text { if } 1 \leq x<4 \\\10-2 x & \text { if } 4 \leq x \leq 6\end{array}\right.$$

Additional integrals Use a change of variables to evaluate the following integrals. $$\int_{1}^{2} \frac{4}{9 x^{2}+6 x+1} d x$$
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