How do you evaluate the definite integral #int 2x-1# from #[0,5]#?

Answer 1

Emilia Aguilar

#20#

#\int_0^5 (2x - 1) dx = [x^2 - x]_0^5 = 25 - 5 = 20#

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Answer 2

Axel Alvarez

To evaluate the definite integral ∫(2x - 1) dx from [0,5], you first need to find the antiderivative of the integrand, which is (x^2 - x). Then, you evaluate this antiderivative at the upper limit of integration (5) and subtract the value of the antiderivative at the lower limit of integration (0).

So,

∫(2x - 1) dx = (x^2 - x) evaluated from 0 to 5.

Substituting the upper limit:

= (5^2 - 5)

Substituting the lower limit:

= (0^2 - 0)

Now, subtract the lower limit from the upper limit:

= (5^2 - 5) - (0^2 - 0)

= (25 - 5) - (0 - 0)

= 20 - 0

= 20.

Therefore, the value of the definite integral ∫(2x - 1) dx from [0,5] is 20.

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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.