F(x) = definite integral of Sqrt(t^2-9)dt on [4,x]. Find: F(4)?

because, when you evaluate a definite integral over the interval [a,b] you take the anti-derivative evaluated at a and subtract from it the anti-derivative evaluated at b.

in your example, the interval is [4,4] which means the integral is taken, evaluated at 4, then evaluated at 4 again and the two (identical values) are subtracted.

in other words, 4(x)- 4(x) = 0, where x = anything.

If this is a homework problem that you're supposed to work on yourself, then asking this question here constitutes academic dishonesty (cheating).

Let's assume it's a simple Riemann (Calculus 101) integral, and that it is being evaluated over a t interval of zero length (from 4 to 4). If the integrand has no singularities in the vicinity of 4, then the integral is well defined (its upper and lower limits at the interval end points are equal) and has value zero. You must check this condition on the integrand because if it isn't met, the integral isn't defined even if the interval endpoints are identical.

The function sqrt(t^2-9) is indeed well defined and continuous for |t| .ge. 3, so the conclusion is as stated.

One way to "see" that the answer is zero is to go back to the definition of the integral (in terms of upper and lower sums of step approximations). For example,

F(x) =(approx) (x-4) f(4) (lower sum)

=(approx) (x-4) sqrt(7)

which converges to zero as x converges to 4.

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If this still doesn't convince you, then by using trig substitution and integration by parts, you can show:

integral[sqrt(t^2 - a^2)]dt

= (t/2)sqrt(t^2 - a^2) - ((a^2)/2)ln (t + sqrt(t^2 - a^2)) + const.

If you're taking a calculus class, it's worthwhile to derive this. Perhaps inability to do this derivation is what you mean by, "I really have no idea why."

It should then be clear that the indefinite integral is well defined in the the vicinity of t=4, and that the definite integral from [4,4] evaluates to zero.

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Before YOU get on a high horse, the proper person to ask your question is your teacher, and if you submit any of the answers given here as your own without proper attribution, then it is cheating plain and simple.

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If the integrand in question is actually sqrt(t^2 + 9), then the solutions I offered are actually even simpler to obtain. The indefinite integral provided most recently needs to be changed, but since you're following this, I'm sure you'll be able to handle this without any difficulty.

The area under the graph from x=4 to x=4 is 0.