Questions about integration

Question

I'm still a bit confused about definite integration although got the basic idea of how to do integration. The problem is to integrate functions on a uniform distribution over [50, 150].

Firstly about the expected value, why the expected value of the uniform distribution over [50, 150]) is the integral of this interval?

Then we get

I understood that the probability over the interval [50, 150] is because its area has to be 1, but why should be ? What is the in this case?

Then I integrate 2 functions with this uniform distribution, and .

For the first function, the answer is . Why we have before the function? And why it disappeared afterwards? I thought according to the integration rules ∫cf(x) dx = c∫f(x) dx but I was wrong, why the 1/100 is not integrated and disappeared?

For the second function I'm even more confused! The answer is , why is kept this time, unlike the first function? Most consufing part is why the integration of becomes ? What I did was wrong:

Can you show me how to solve it?

Answer 1

It seems your fundamental difficulty is with definitions of probability concepts, not so much with the process of integration.

If $X \sim Unif(50, 150),$ then the density function $f_X(x)$ has three parts:

(i) For $x < 50,\;f_X(x) = 0;\;$;

(ii) for $50 \le x \le 150,\;f_X(x) = 1/100 = 0.01;$ and

(iii) for $x > 150,\;f_X(x) = 0.$

If you want to find $P(X \le 75),$ you use the density function as follows:

$$P(X \le 75) = \int_{-\infty}^{75} f_X(x)\,dx = \int_{-\infty}^{50} 0\,dx + \int_{50}^{75} 0.01\,dx = ) = 0 +[.01x]_{50}^{75} = .25.$$

Actually, the uniform density function is so simple that you don't need calculus to find this probability: Draw a rectangle above the interval $[50,150]$ that is 0.01 high. You can see that 1/4 of its area lies between 50 and 75 (and no area below 50).

Notice that the only way $x$ appears in the density function of $Unif(50,150)$ is in defining the intervals. On the interval $(50,150),$ the density function is simply "0.01," not "0.01x".

The definition of the expectation of a continuous random variable $X$ with density function $f_X(x)$ is defined as $$E(X) = \int_{-\infty}^{\infty} xf_X(x)\,dx,$$ provided that the 'improper' integral converges absolutely. (But the proviso is not an issue for a uniform distribution.) Notice that this is not what you have written in your question.

Your integration to find $E(X)$ is correct, but your interpretation of the details of it is not. Here is the full computation.

$$E(X) = \int_{-\infty}^{\infty} x\,f_X(x)\,dx = \int_{-\infty}^{50} x(0)\,dx + \int_{50}^{75} x(0.01)\,dx +\int_{150}^{\infty} x(0)\,dx\\ = 0 + \int_{50}^{75} 0.01x\,dx + 0 = \cdots = 100.$$

The density function $f_X(x)$ is a factor in the integrand; the entire integrand is $xf_X(x).$

Sometimes, geometrically speaking, it is useful to regard $E(X)$ as the 'balance point' of the density function. If you look at the rectangle you drew to represent $f_X(x),$ you will see that it would 'balance' if you put a fulcrum at $x = 100$ on the x-axis.

In the rest of your question, you seem to be trying to find the expectation of the random variable $e^{.01X}.$ But from what you have written, I can't be sure. Maybe edit the last part to show the question clearly, then a revision of your attempted answer following my method for $E(X)$. If that is what you are trying to do, then start with $E(e^{.01X}) = \int_{-\infty}^{\infty} e^{.01x}f_X(x)\,dx = \cdots.$ Then I (or someone else) can check your answer.

Addendum: In case you are interested in looking at approximate answers from a simulation in R, some are shown below. (If you have Matlab, or similar software, available and know how to use it, you should be able to get exact answers, but that is another discussion.)

 m = 10^6; x = runif(m, 50, 150)
 mean(x < 75)  
 ## 0.249902  # Approx. P(X < 75) = 1/4
 mean(x)
 ## 100.0025  # Approx. E(X) = 100
 mean(exp(.01*x))
 ## 2.832915
 mean(1-exp(-.01*x))
 ## 0.6166262