Q 16 Suppose f is an integrable funct... [FREE SOLUTION]

Chapter 4: Q 16 (page 353)

Suppose f is an integrable function [a, b] and k is a
real number. Use pictures of Riemann sums to illustrate
that the right sum for the function kf (x) on [a, b] is k
times the value of the right sum (with the same n) for
f on [a, b]. What happens as n鈫掆垶? What does this
exercise say about the definite integrals
${鈭�}_{a}^{b} f (x) d x a n d {鈭�}_{a}^{b} k f (x) d x ?$

Short Answer

Expert verified

For two randomly chosen functions, derive (1) and (2), f(x)=x+2 and $g (x) = e^{- x}$ we have,

鈭� \frac{f (x)}{g (x)} d x 鈮� \frac{鈭� f (x) d x}{鈭� g (x) d x} .

Step by step solution

Step 1: Given information: y

We consider two functions, f(x) and g(x), which have the following properties:

$鈭� \frac{f (x)}{g (x)} d x 鈮� \frac{鈭� f (x) d x}{鈭� g (x) d x}$

Method/Formula: We arbitrarily consider two functions, f(x) and g(x).

Calculation:

Suppose f(x)=x+2 and g(x)=e^-x

Therefore, $鈭� \frac{f (x)}{g (x)} d x = 鈭� \frac{x + 2}{e^{- x}} d x = 鈭� (x e^{x} + 2 e^{x}) d x = 鈭� x e^{x} d x + 鈭� 2 e^{x} d x$

$鈭� \frac{f (x)}{g (x)} d x = x 鈭� e^{x} d x - 鈭� (\frac{d}{d x} x 鈭� e^{x} d x) + 鈭� 2 e^{x} d x$

$鈭� \frac{f (x)}{g (x)} d x = x e^{x} - 鈭� (1 路鈭� e^{x} d x) + 2 e^{x}$

$鈭� \frac{f (x)}{g (x)} d x = x e^{x} - e^{x} + 2 e^{x} = x e^{x} + e^{x}$

Thus for elected functions f and g we have,

$鈭� \frac{f (x)}{g (x)} d x = x e^{x} + e^{x}$ ..... (1)

Now, $鈭� f (x) d x = 鈭� (x + 2) d x = 鈭� x d x + 鈭� 2 d x = \frac{x^{2}}{2} + 2 x$ and $鈭� g (x) d x = 鈭� e^{- x} d x = \frac{e^{- x}}{- 1} = e^{- x}$

Therefore, $\frac{鈭� f (x) d x}{鈭� g (x) d x} = \frac{\frac{x^{2}}{2} + 2 x}{e^{x}} = \frac{x^{2} + 4 x}{- 2 e^{- x}} = - \frac{1}{2} (x^{2} + 4 x) e^{x}$ ... (2)

From results (1) an (2) for two arbitrarily chosen functions f(x) = x + 2 and g(x) = e^-x we have,

Hence proved

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Short Answer

Step by step solution

Step 1: Given information: y

Method/Formula: We arbitrarily consider two functions, f(x) and g(x).

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