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Chapter 10: Q. 9 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.
role="math" localid="1654760569311" $f (x) = \frac{1}{{(2 鈭� 5 x)}^{3}}$

Short Answer

Expert verified

The final integrate function is $鈭� \frac{1}{5} 鈭� \frac{1}{{(t)}^{3}} dt = \frac{1}{10 {(2 鈭� 5 x)}^{2}} + C$

Step by step solution

01

Step 1. Given Information.

The given function is $f (x) = \frac{1}{{(2 鈭� 5 x)}^{3}}$ .

02

Step 2. Find the integration of the given function.

The given function is $f (x) = \frac{1}{{(2 鈭� 5 x)}^{3}} . ... (i)$

$F' (x) = f (x)$

Adding up the numbers (1).

localid="1654761391962" $鈭� \frac{1}{(2 鈭� 5 x)^{3}} d x$

Multiplying and dividing by 5.

$\frac{1}{5} 鈭� \frac{5}{(2 鈭� 5 x)^{3}} dx . . . (ii)$

Assume that $2 鈭� 5 x = t$ distinguishing in terms of x.

$5 xdx = 鈭� dt$

Replace the value in equation (2).

$鈭� \frac{1}{5} 鈭� \frac{1}{{(t)}^{3}} dt$

Now integrate the obtained function.

localid="1654761744441" $\begin{array}{l} 鈭� \frac{1}{5} 鈭� \frac{1}{{(t)}^{3}} dt = 鈭� \frac{1}{5} 鈭� {(t)}^{鈭� 3} dt \\ 鈭� \frac{1}{5} 鈭� \frac{1}{{(t)}^{3}} dt = 鈭� \frac{1}{5} \frac{{[t]}^{鈭� 3 + 1}}{鈭� 3 + 1} + C \\ 鈭� \frac{1}{5} 鈭� \frac{1}{{(t)}^{3}} dt = \frac{1}{5} 脳 (\frac{1}{2}) {[t]}^{- 2} + C \\ 鈭� \frac{1}{5} 鈭� \frac{1}{{(t)}^{3}} dt = \frac{1}{10 {(t)}^{2}} + C \end{array}$

C denotes the integration constant.

Adding the value of t to the equation above.

$鈭� \frac{1}{5} 鈭� \frac{1}{{(t)}^{3}} dt = \frac{1}{10 {(2 鈭� 5 x)}^{2}} + C$

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