One
This one I know that it will be right.
[tex] \int\limits^4_0 {x^3} \, dx= \frac{x^4}{4}|\limits^4_0=\frac{256}{4} -0=64 [/tex]
Two
The answer to this one is either 2 or 4. The way I'm reading the question, y =0 or the x axis is the lower limit. I don't think you include anything where y < 0
I can't get the latex to work properly. But here is the way I would like it to be
Integral of Cos(x) dx = sin(x) with limits between pi/2 and 0
sin(pi/2) - sin(0) = 1 - 0 = 1
The other point where this is above the x axis begins at 3/2pi and ends at 2p
so you have to take sin(2pi) - (sin(3/2 pi) = 0 - - 1 = 1
The total area above the x axis is 1 + 1 = 2 <<<<<<<< answer
If you include the area between pi/2 and 3/2 pi it would be
sin(pi/2) - sin(3/2 pi) = 1 - -1 = 2 Which added to the previous answer will give 4. You have to try them both. If you put 2 in and it is wrong then use 4.
Three
The problem with 1/x is that it is undefined even approaching the lower limit it is very good. My first instinct is to tell you that if can't be done. Or that it shouldn't be done. Try 1*10^-6 as your lower limit.
integrate (1/x) with the limits of 1 and 1/1000000
ln(x) with limits between 1 and 1/1000000
ln(1) - ln(1/1000000)
0 - (-13.815) = 13.816
1 millionth is no where near 0. It's just enough to give you an idea that since the graph is returning something very large, zero will give you something that is infinite I've included the graph of this function to show you why.
Five
bounded Below by the graph of y = x - 2
bounded Above by the graph of y = sqrt(x)
I've made a graph of this one as well. The only way you can do this is by integrating the sqrt(x) and subtracting the area of the right triangle with a base going from 2 to 4
y = integral sqrt(x) between 4 and 0
y = 2/3 x^3/2 between 4 and 0
y = 2/3 * 8 - 0
y = 16/3
Now what you must do is subtract the area of area of the triangle
b = 4 - 2
b = 2
h = 2 - 0
h = 2
Area = 1/2 b * h
Area = 1/2 * 2*2
Area = 2
Total area = 16/3 - 2 = 32/6 - 12/6 = 20/6 <<<<< Answer
Final Answer.
Number 4 is all yours. It's done mostly like this one except you will need two integrals.
