Finding the exact value of an integral without calculator

[rubenhero]

Homework Statement

Find the exact value of: ∫_-5⁵ √25-x² dx without using your calculator. (Hint: Consider the geometric significance of the definite integral.)

Homework Equations

integrate and find antiderivative and evaluate at A and B.

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!

 

[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 learned about using substitutions to solve integrals yet?

 

[rubenhero]

thank you for your quick response,

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 learned about using substitutions to solve integrals yet?

is $y=\sqrt{1-x^2}$ a circle?, oh a circle isn't continuous so i can't 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)u^3/2, 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?

 

[HallsofIvy]

thank you for your quick response,


is $y=\sqrt{1-x^2}$ a circle?, oh a circle isn't continuous so i can't do antiderivatives?

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.

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)u^3/2, 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?

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

 

[rubenhero]

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