To determine

To estimate:

The amount of water in the tank four days later by using the Midpoint Rule.

Answer

28,320 L

Explanation

1) Concept:

Use the Net Change theorem and the Midpoint Rule.

2) Theorem:

The integral of a rate of change is the net change.

∫abF'(x)dx=Fb-F(a)

3) Given:

The amount of water in the tank at t=0 is 25,000 L.

4) Calculation:

Initially, 25,000 L water is present in the tank.

Therefore, by the Net Change theorem, the amount of water in the tank four days later is given by

=25000+∫04r(t)dt

Find ∫04r(t)dt by using the Midpoint Rule.

With n=4, a=0 and b=4 interval width is

∆t=b-an

Substitute values.

∆t=4-04

∆t=1

And the midpoints are

t1-=12t0+t1=120+1=121=0.5

t2-=12t1+t2=121+2=123=1.5

t3-=12t2+t3=122+3=125=2.5

t4-=12t3+t4=123+4=127=3.5

So, the value of integral is

∫04r(t)dt≈M4=∑i=14r(ti-)∆t

=∆x(rt1-+rt2-+rt3-+rt4-)

=∆x(r0.5+r1.5+r2.5+r(3.5))

From the given graph,

r0.5=1500

r1.5=1770

r2.5=740

r3.5=-690

Substitute these values in Riemann sum.

M4=1(1500+1770+740-690)

Simplify.

M4=13320=3,320

Substitute in the amount of water in the tank four days later.

=25,000+3,320

=28,320

Therefore, the amount of water in the tank four days later is 28,320 L.

Conclusion:

Therefore, the amount of water in the tank four days later is 28,320 L.