MHB What is the solution to the integral without an equation?

  • Thread starter Thread starter karush
  • Start date Start date
  • Tags Tags
    Integral
karush
Gold Member
MHB
Messages
3,240
Reaction score
5
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
 
Last edited:
Physics news on Phys.org
(Wave)

$$\int_0^4 (2-p(x)) dx= \int_0^4 2 dx - \int_0^4 p(x) dx= 2 \cdot 4-6=2$$
 
How do determine what the area 0-2 and 2-4 can't assume area is half?
For iii
 
Last edited:
karush said:
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 $\int_a^c f(x) dx+ \int_c^b f(x) dx= \int_a^b f(x) dx$.
 
$\int_{0}^{2}p\left(x\right) \,dx
+\int_{2}^{4}p\left(x\right) \,dx
-\int_{0}^{2}2 \,dx
+\int_{0}^{2}1 \,dx$

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

$6-4+2=4$
 
Last edited:

Similar threads

Back
Top