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How do you evaluate the definite integral #int 4x-5# from #[1,2]#?

Answer 1

Emily Ammerman

#1#.

Recall the Fundamental Principle of Definite Integration :

#if intf(x)dx=F(x)+C, then, int_a^bf(x)dx=F(b)-F(a)=[F(x)]_a^b#.

#int(4x-5)dx=4x^2/2-5x+C=2x^2-5x+C#

#:. int_1^2(4x-5)dx=[2x^2-5x]_1^2=[(8-10)-(2-5)]=-2+3=1#.

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Answer 2

Charles Anctil

To evaluate the definite integral ∫(4x - 5) dx from 1 to 2, you first find the antiderivative of the integrand, which is (2x^2 - 5x). Then, you evaluate this antiderivative at the upper and lower limits of integration and subtract the value at the lower limit from the value at the upper limit. So, you have:

[ (2(2)^2 - 5(2)) ] - [ (2(1)^2 - 5(1)) ] = [ (8 - 10) ] - [ (2 - 5) ] = [ (8 - 10) ] - [ -3 ] = -2 + 3 = 1

Therefore, the value of the definite integral ∫(4x - 5) dx from 1 to 2 is 1.

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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.