Question

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→∞? What does this

exercise say about the definite integrals

${\int }_a^{b}f\left(x\right)dxand{\int }_a^{b}kf\left(x\right)dx?$

Short answer

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

$\int \frac{f\left(x\right)}{g\left(x\right)}dx\ne \frac{\int f\left(x\right)dx}{\int g\left(x\right)dx}.$

Step-by-step solution

Step 1: Given information: y

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

$\int \frac{f\left(x\right)}{g\left(x\right)}dx\ne \frac{\int f\left(x\right)dx}{\int g\left(x\right)dx}$

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

Calculation:

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

Therefore, $\int \frac{f\left(x\right)}{g\left(x\right)}dx=\int \frac{x+2}{e^{-x}}dx=\int \left(x{e}^x+2e^x\right)dx=\int x{e}^{x}dx+\int 2e^{x}dx$

$\int \frac{f\left(x\right)}{g\left(x\right)}dx=x\int e^{x}dx-\int \left(\frac{d}{dx}x\int e^{x}dx\right)+\int 2e^{x}dx$

$\int \frac{f\left(x\right)}{g\left(x\right)}dx=x{e}^x-\int \left(1·\int e^{x}dx\right)+2e^x$

$\int \frac{f\left(x\right)}{g\left(x\right)}dx=x{e}^x-e^x+2e^x=x{e}^x+e^x$

Thus for elected functions f and g we have,

$\int \frac{f\left(x\right)}{g\left(x\right)}dx=x{e}^x+e^x$..... (1)

Now, $\int f\left(x\right)dx=\int (x+2)dx=\int xdx+\int 2dx=\frac{x^2}{2}+2x$and $\int g\left(x\right)dx=\int e^{-x}dx=\frac{e^{-x}}{-1}=e^{-x}$

Therefore, $\frac{\int f\left(x\right)dx}{\int g\left(x\right)dx}=\frac{\frac{x^2}{2}+2x}{e^x}=\frac{x^2+4x}{-2e^{-x}}=-\frac{1}{2}\left(x^2+4x\right)e^x$... (2)

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

Hence proved