#### 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.