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How do you evaluate the definite integral #int (4t^3-2t)dt# from [-1,1]?

Answer 1

Evelyn Agard

#0.#

At a glance, we can answer #0#, if we use the following Result :

If,#f:[-a,a] to RR" is continuous odd function, "int_(-a)^a f(x)dx=0.#

Otherwise,

#int_(-1)^1 (4t^3-2t)dt=2[(2t^4)/4-t^2/2]_-1^1#

#=2[(2/4-1/2)-(2/4-1/2)]#

#=0#.

Enjoy Maths.!

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Answer 2

Charles Arocha

To evaluate the definite integral (\int_{-1}^{1} (4t^3 - 2t) dt), you first find the antiderivative of the integrand, then evaluate it at the upper and lower limits of integration, and finally subtract the result at the lower limit from the result at the upper limit.

The antiderivative of (4t^3 - 2t) is (t^4 - t^2).

Evaluating this antiderivative at the upper limit (1) yields (1^4 - 1^2 = 1 - 1 = 0).

Evaluating at the lower limit (-1) yields ((-1)^4 - (-1)^2 = 1 - 1 = 0).

Therefore, the definite integral is (0 - 0 = 0).

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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.