Find the distance from 0 to 3 of the ∫ l3t-5ldt. I don't understand how the answer is 41/6.
∫ l3t-5ldt from 0 to 3
assuming it means absolute value, then
= | [ (3/2t^2 - 5t ] | from 0 to 3
= | (3/2)(9) - 5(3) - 0 |
= 3/2
not sure what you mean by the "distance"
If you mean the value of the integral, note that the vertex of the graph is at t = 5/3, where 3t-5=0
So, the integral is
∫[0,5/3] -(3t-5) dt + ∫[5/3,3] (3t-5)dt
The area is just the sum of the two triangles.
see
www.wolframalpha.com/input/?i=integral%5B0..3%5D+%7C3x-5%7C+dx
To find the distance from 0 to 3 of the integral ∫ (3t - 5) dt, we need to first evaluate the integral and then take the absolute value of the result.
To evaluate the integral, we can apply the power rule of integration. The power rule states that if we have a term of the form ax^n, the integral is (a/(n+1))x^(n+1) + C.
In our case, we have the integrand l3t - 5l dt. This can be split into two separate integrals, one for each term:
∫ (3t) dt - ∫ (5) dt.
Applying the power rule for integrals, we get:
(3/2)t^2 - 5t + C.
To evaluate the integral over the interval from 0 to 3, we need to substitute the bounds into the antiderivative and subtract the result for the lower bound from the result for the upper bound.
Substituting the upper bound t = 3 into the antiderivative:
(3/2)(3)^2 - 5(3) + C,
And substituting the lower bound t = 0 into the antiderivative:
(3/2)(0)^2 - 5(0) + C.
Simplifying these expressions gives us:
(3/2)(9) - 15 + C, and
(3/2)(0) - 0 + C.
Next, we subtract the result for the lower bound from the result for the upper bound:
[(3/2)(9) - 15 + C] - [(3/2)(0) - 0 + C].
Simplifying further:
(27/2) - 15 = 27/2 - 30/2 = -3/2.
Finally, we take the absolute value of the result to find the distance, which gives us:
|-3/2| = 3/2.
So, it seems like there may have been an error in the calculation or in the given answer. The correct distance from 0 to 3 of the ∫ (3t - 5) dt is 3/2, not 41/6.