Let F(x)=∫0x (t−1) dt. Find a formula for F and its derivative, and explain where F decreases and why F(2)=0.
F(x)=∫0x(t−1)dt
1
Find an antiderivative of the integrand
Part 2 needs any antiderivative of t−1. The function t2/2−t has derivative t−1, so it qualifies. Call it G to keep it apart from F.
G(t)=2t2−t
2
Subtract the endpoint values
By Part 2, F(x)=G(x)−G(0), the upper limit minus the lower one. G(0)=0, so nothing is subtracted.
F(x)=2x2−x
3
Read the slope
Part 1 says F′(x) is the integrand with x in place of t, so F′(x)=x−1. Differentiating the formula x2/2−x gives the same x−1, a check on the antiderivative.
F′(x)=x−1
4
Where F decreases
F′(x)=x−1 is negative for x<1. On [0,1] the integrand is below the axis, so each new strip subtracts area and F falls, reaching −1/2 at x=1. After that the strips add.
5
Why F(2)=0
From 0 to 1 the signed area is −1/2, a triangle below the axis. From 1 to 2 it is +1/2, a matching triangle above it. They cancel, so F(2)=0 even though the total area is 1.
F(2)=0
Result
F(x)=x2/2−x and F′(x)=x−1.F decreases for x<1 and increases for x>1, and F(2)=0 because the areas below and above the axis cancel.
Your turn
1. One like the example
Let F(x)=∫2x (2t−4) dt. Find a formula for F, then F′(x) and F(4).
Show the answer and explanation
F(x)=x2−4x+4,F′(x)=2x−4 and F(4)=4.
An antiderivative of 2t−4 is t2−4t. By Part 2, F(x)=(x2−4x)−(4−8)=x2−4x+4. Part 1 gives F′(x)=2x−4 straight from the integrand. F(4)=16−16+4=4, the area of the triangle under 2t−4 from 2to 4.
F(x)=∫2x(2t−4)dtF(x)=x2−4x+4F′(x)=2x−4F(4)=4
2. A variation
Let H(x)=∫1x3 (1/t) dt for x>0. Find H′(x). This time the upper limit is x3, not x.
Show the answer and explanation
H′(x)=3/x.
Part 1 with the chain rule puts the upper limit into the integrand, giving 1/x3, and multiplies by the upper limit’s derivative, 3x2. The product 3x2/x3 is 3/x. Integrating first agrees. H(x)=ln(x3)−ln1=ln(x3)=3lnx, whose derivative is 3/x.
A reservoir gains water at 30−6t thousand liters per hour, t hours after noon. A negative rate means it is losing water. Let W(t)=∫0t (30−6s) ds be the water gained since noon. When is W largest, and how large is it?
Show the answer and explanation
At t=5, which is 5 p.m., when W(5)=75 thousand liters.
By Part 1, W′(t)=30−6t. It is positive before t=5 and negative after, so W rises and then falls, with its largest value at t=5. An antiderivative gives W(t)=30t−3t2, and W(5)=150−75=75.
W(t)=∫0t(30−6s)dsW(t)=30t−3t2W′(t)=30−6tW(5)=75
Keep exploring
In Accumulation / FTC, move x from 0to 2.F falls to −1/2 at x=1, where its tangent is flat, then climbs back to 0 at x=2.
