Home > Calculus > Formal Definition of the Definite Integral

How do you find the definite integral for: #(sqrt(b^2-a^2)) da# for the intervals #[0, b]#?

Answer 1

Used the geometric meaning of the definite integral.

The graph of #y = sqrt(r^2-x^2)# is the upper semicircle centered at #(0,0)# with radius #r#

The integral from #0# to #r# is, therefore one fourth of the area of the circle with radius #r#.

In this problem, we have changed the independent variable to #a# and the name of the radius to #b#, but the reasoning is unchanged.

#int_0^b sqrt(b^2-a^2) da = 1/4(pi b^2)#

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Answer 2

Cameron Alexander

To find the definite integral of sqrt(b^2 - a^2) with respect to 'a' over the interval [0, b], you can use trigonometric substitution. Let 'a' = b * sin(theta), then da = b * cos(theta) d(theta). Substituting these into the integral yields:

∫sqrt(b^2 - a^2) da = ∫sqrt(b^2 - (b * sin(theta))^2) * b * cos(theta) d(theta)

Simplify this expression and integrate it with respect to theta. After integrating, substitute back the original variable 'a' and evaluate the integral over the interval [0, b].

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.