What is the solution to the integral without an equation?

[karush]

https://www.physicsforums.com/attachments/5154http://mathhelpboards.com/attachment.php?attachmentid=5153&stc=1
I didn't see how they got ii and iii
Bk answers are in red

 

[evinda]

(Wave)

$$\int_0^4 (2-p(x)) dx= \int_0^4 2 dx - \int_0^4 p(x) dx= 2 \cdot 4-6=2$$

 

[karush]

How do determine what the area 0-2 and 2-4 can't assume area is half?
For iii

 

[evinda]

How do determine what the area 0-2 and 2-4 can't assume area is half?
For iii

What do you mean? I haven't understood your question.In general it holds that $\int_a^c f(x) dx+ \int_c^b f(x) dx= \int_a^b f(x) dx$.

 

[karush]

$\int_{0}^{2}p\left(x\right) \,dx
+\int_{2}^{4}p\left(x\right) \,dx
-\int_{0}^{2}2 \,dx
+\int_{0}^{2}1 \,dx$

$\int_{0}^{4}p(x) \,dx-4+2$

$6-4+2=4$