# Finding the exact value of an integral without calculator

1. May 11, 2012

### rubenhero

1. The problem statement, all variables and given/known data
Find the exact value of: ∫-55 √25-x2 dx without using your calculator. (Hint: Consider the geometric significance of the definite integral.)

2. Relevant equations
integrate and find antiderivative and evaluate at A and B.

3. The attempt at a solution
-55 √25-x2 dx
=[2/3(25-x2)3/2*2x]-55
=(2/3(25-52)3/2*2*5)-(2/3(25--52)3/2*2*-5)
=0-0

I don't think I got the right answer, I thought i always find the antiderivative to integrate.
Is there something I am doing wrong? Any help would be appreciated!

2. May 11, 2012

### Mentallic

Yes it's definitely wrong, and also it's very hard to follow what you did to tell you where you went wrong.

What does the graph $y=\sqrt{1-x^2}$ represent?

And have you learnt about using substitutions to solve integrals yet?

3. May 11, 2012

### rubenhero

thank you for your quick response,
is $y=\sqrt{1-x^2}$ a circle?, oh a circle isn't continuous so i cant do antiderivatives?
i have learned substitutions but if i let u be 25-x2, du = -2x dx, i would have to put a (-1/2x) in front of the integral, then integrate u to be (2/3)u3/2, wouldn't i still end up with (25-(52))-(25-(-5)2))
would i just use the area of a circle formula ∏r2 for this problem?

4. May 11, 2012

### HallsofIvy

Staff Emeritus
A circle would be $x^2+ y^2= 1$- y can be both positive and negative. $y=\sqrt{1- x^2}$ requires that y be non-negative- it is a semi-circle.

Yes, that is the whole point of hint- the area under the semi-circle is half the area of a circle of radius 1.

5. May 11, 2012

### rubenhero

thank you, that makes sense now. the radius is 5 so it would be (1/2)∏52.