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Chapter 6: Q. 52 (page 499)

Each definite integral represents the arc length of a function f(x) on an interval [a, b]. Determine the function and interval.
${鈭�}_{0}^{1} \sqrt{1 + 9 x^{4}} d x$

Short Answer

Expert verified

The function is $f (x) = x^{3}$ and interval is $[a, b] = [0, 1]$ .

Step by step solution

01

Step 1. Given information.

Consider the given function is ${鈭�}_{0}^{1} \sqrt{1 + 9 x^{4}} d x$ .

02

Step 2. Use arc length formula.

The formula for a function to find the arc length from $x = a$ to $x = b$ is given by localid="1649291498730" role="math" ${鈭�}_{a}^{b} \sqrt{1 + {(f^{'} (x))}^{2}} d x$ .

03

Step 3. Find derivative of given function.

Compare given definite integral function with the arc length formula.

$\begin{array}{rcl} {鈭�}_{a}^{b} \sqrt{1 + {(f^{'} (x))}^{2}} d x & = & {鈭�}_{0}^{1} \sqrt{1 + 9 x^{4}} d x \\ = & {鈭�}_{0}^{1} \sqrt{1 + {(3 x^{2})}^{2}} d x \end{array}$

Therefore the derivative function will be $f^{'} (x) = 3 x^{2}$ and the interval is $[a, b] = [0, 1]$ .

04

Step 4. Find function fx.

The function whose differentiation is $3 x^{2}$ is $f (x) = x^{3}$ .

Thus the required function is $f (x) = x^{3}$ and $[a, b] = [0, 1]$ .

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Most popular questions from this chapter

The centroid of the region between the graphs of $f (x) = \sqrt{x}$ and $g (x) = 4 - x$ on $[0, 4]$

Use antidifferentiation and/or separation of variables to solve each of the initial-value problems in Exercises 29-52.

$\frac{d y}{d x} = x y + x + y + 1, y (0) = c$

$\frac{d y}{d x} = 3 - 4 y, y (0) = 2$

Use antidifferentiation and/or separation of variables to solve each of the initial-value problems in Exercises 29鈥�52.

30. $\frac{d y}{d x} = 3 x, y (0) = 4$

Solve the initial-value problem

\frac{d y}{d x} = 2 y (1 - 5 y), y (0) = 1

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