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Chapter 10: Q. 7 TF (page 777)

Finding antiderivatives by undoing the chain rule: For each function f that follows, find a function F with the property that $F (x) = f (x)$ . You may have to guess and check to find such a function
$f (x) = x \sqrt{1 + x^{2}}$

Short Answer

Expert verified

The Final Answer is $\frac{1}{2} 鈭� \sqrt{t} dt = \frac{1}{3} {[1 + x^{2}]}^{\frac{3}{2}} + C$

Step by step solution

01

Given information

The given function is $f (x) = x \sqrt{1 + x^{2}}$ .

02

Calculations

The given function is $f (x) = x \sqrt{1 + x^{2}} . . . (i)$

$F^{鈥�} (x) = f (x)$

Adding up the numbers (1)

$鈭� x \sqrt{1 + x^{2}} dx$

Multiplying and dividing by 2.

$\frac{1}{2} 鈭� 2 x \sqrt{1 + x^{2}} dx . . . (ii)$

Assume that, $1 + x^{2} = t$ distinguishing in terms of x.

$2 xdx = dt$

Replace the value in equation (2).

$\frac{1}{2} 鈭� \sqrt{t} dt$

Now, integrate the obtained function.

\begin{aligned} \frac{1}{2} 鈭� \sqrt{t} dt = \frac{1}{2} \frac{| t |^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} + C \\ \frac{1}{2} 鈭� \sqrt{t} dt = \frac{1}{2} 脳 (\frac{2}{3}) [t]^{\frac{3}{2}} + C \\ \frac{1}{2} 鈭� \sqrt{t} dt = \frac{1}{3} [t]^{\frac{3}{2}} + C \end{aligned}

Where C denotes the integration constant.

Adding the value of t to the equation above.

$\frac{1}{2} 鈭� \sqrt{t} dt = \frac{1}{3} {[1 + x^{2}]}^{\frac{3}{2}} + C$

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Most popular questions from this chapter

Find a vector in the direction of $<- 1, 2, 3>$ with magnitude 3.

Find a vector in the direction of $<9, 0, - 6>$ and with magnitude 7.

Determine whether each of the statements that follow is true or false. If a statement is true, explain why. If a statement is false, provide a counterexample.

(a) True or False: The sum formulas in Theorem 4.4 can be applied only to sums whose starting index value is $k = 1$ .

(b) True or False: ${鈭�}_{k = 0}^{n} 鈥� \frac{1}{k + 1} + {鈭�}_{k = 1}^{n} 鈥� k^{2}$ is equal to ${鈭�}_{k = 0}^{n} 鈥� \frac{k^{3} + k^{2} + 1}{k + 1}$ .

(c) True or False: ${鈭�}_{k = 1}^{n} 鈥� \frac{1}{k + 1} + {鈭�}_{k = 0}^{n} 鈥� k^{2}$ is equal to ${鈭�}_{k = 1}^{n} 鈥� \frac{k^{3} + k^{2} + 1}{k + 1}$ .

(d) True or False: $({鈭�}_{k = 1}^{n} 鈥� \frac{1}{k + 1}) ({鈭�}_{k = 1}^{n} 鈥� k^{2})$ is equal to ${鈭�}_{k = 1}^{n} 鈥� \frac{k^{2}}{k + 1}$ .

(e) True or False: ${鈭�}_{k = 0}^{m} 鈥� \sqrt{k} + {鈭�}_{k = m}^{n} 鈥� \sqrt{k}$ is equal to ${鈭�}_{k = 0}^{n} 鈥� \sqrt{k}$ .

(f) True or False: ${鈭�}_{k = 0}^{n} 鈥� a_{k} = 鈭� a_{0} 鈭� a_{n} + {鈭�}_{k = 1}^{n 鈭� 1} 鈥� a_{k}$ .

(g) True or False: ${({鈭�}_{k = 1}^{10} 鈥� a_{k})}^{2} = {鈭�}_{k = 1}^{10} 鈥� a_{k}^{2}$ .

(h) True or False: ${鈭�}_{k = 1}^{n} 鈥� {(e^{x})}^{2} = \frac{e^{x} (e^{x} + 1) (2 e^{x} + 1)}{6}$ .

Consider the function f shown in the graph next at the right. Use the graph to make a rough estimate of the average value of f on [鈭�4, 4], and illustrate this average value as a height on the graph.

Find a vector of length 3 that points in the direction opposite to $<1, - 2, 3>$ .

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