# Understanding expected value of loss

I am having a trouble understanding a subtle difference in the answer of two problems.

Problem one:

An insurance policy reimburses a loss up to a benefit limit of 10. The policyholder's loss, Y. follows a distribution with density function:

$f(x) = \begin{cases} \frac{2}{x^3}, & \text{if$x \gt 1$} \\ 0, & \text{otherwise} \end{cases}$

What is the expected value of the benefit paid under the insurance policy?

Problem two:

An auto insurance company insures an automobile woth 15,000 for one year under a policy with a 1000 deductible. During the policy year there is a .04 chance of partial damage to the car and .02 chance of total loss of the car. If there is partial damage to the car, The amount $X$ of damage in the thousands follows a distribution with density function:

$f(x) = \begin{cases} .5003e^{\frac{-x}{2}}, & \text{for$0 \lt x \lt 15$} \\ 0, & \text{otherwise} \end{cases}$

In problem two, Inherently if the policy covers a $15000$ car and the deductible is $1000$ then the maximum loss or payout would be $14,000$ so why when we calculate the answer is it defined as $(0)(.94)+(.04)(.5003)\int_1^{15} e^{\frac{-x}{2}}+(.02)(14)$. The part that I am having the biggest issue with is the last part $(.02)(14)$ Which is contradictory to the first problem where:

$\int_1^{10} y\frac{2}{y^2}dy + \int_{10}^{\infty} 10 \frac{2}{y^2}dy$

The probability of the maximum payout is multiplied by the pdf (which is the likely hood of a value y occurring).

So why is 14 not multiplied by the pdf and integrated in problem two?

It is.   The integral of the pdf over the entire support equals $1$, by definition, and so that is by what it is multiplied.   Also recall the pdf is zero outside the support.
$$\int_0^{15} 14\cdot(0.20)\cdot 0.5003 \mathsf e^{-x/2}~\mathsf d x +\int_{15}^\infty 14\cdot(0.20)\cdot 0~\mathsf d x ~=~ 14\cdot(0.20)$$
PS: also the answer should be : $(0)(.94)+(.04)(.5003)\int_0^{15}x\mathsf e^{-x/2}\mathsf d x+(14)(0.20)$