2. consider the integral z 6 2 1 t 2 dt (a) a. write down—but do not evaluate—the expressions that approximate the integral as a left-sum and as a right sum using n = 2 rectanglesb. Without evaluating either expression, do you think that the left-sum will be an overestimate or understimate of the true are under the curve? How about for the right-sum?c. Evaluate those sums using a calculatord. Repeat the above steps with n = 4 rectangles.

Answers

Answer 1

a) The left-sum approximation for n=2 rectangles is:[tex](1/2)[(2^2)+(1^2)][/tex] and the right-sum approximation is:[tex](1/2)[(1^2)+(0^2)][/tex]

b) The left-sum will be an underestimate of the true area under the curve, while the right-sum will be an overestimate.

c) Evaluating the left-sum approximation gives 1.5, while the right-sum approximation gives 0.5.

d) The left-sum approximation for n=4 rectangles is:[tex](1/4)[(2^2)+(5/4)^2+(1^2)+(1/4)^2],[/tex] and the right-sum approximation is: [tex](1/4)[(1/4)^2+(1/2)^2+(3/4)^2+(1^2)].[/tex]

(a) The integral is:

[tex]\int (from 1 to 2) t^2 dt[/tex]

(b) Using n = 2 rectangles, the width of each rectangle is:

Δt = (2 - 1) / 2 = 0.5

The left-sum approximation is:

[tex]f(1)\Delta t + f(1.5)\Delta t = 1^2(0.5) + 1.5^2(0.5) = 1.25[/tex]

The right-sum approximation is:

[tex]f(1.5)\Delta t + f(2)\Deltat = 1.5^2(0.5) + 2^2(0.5) = 2.25[/tex]

(c) For the left-sum, the rectangles extend from the left side of each interval, so they will underestimate the area under the curve.

For the right-sum, the rectangles extend from the right side of each interval, so they will overestimate the area under the curve.

Using a calculator, we get:

∫(from 1 to 2) t^2 dt ≈ 7/3 = 2.3333

So the left-sum approximation is an underestimate, and the right-sum approximation is an overestimate.

(d) Using n = 4 rectangles, the width of each rectangle is:

Δt = (2 - 1) / 4 = 0.25

The left-sum approximation is:

[tex]f(1)\Delta t + f(1.25)\Delta t + f(1.5)\Delta t + f(1.75)\Delta t = 1^2(0.25) + 1.25^2(0.25) + 1.5^2(0.25) + 1.75^2(0.25) = 1.5625[/tex]The right-sum approximation is:

[tex]f(1.25)\Delta t + f(1.5)\Delta t + f(1.75)\Delta t + f(2)Δt = 1.25^2(0.25) + 1.5^2(0.25) + 1.75^2(0.25) + 2^2(0.25) = 2.0625.[/tex]

Using a calculator, we get:

[tex]\int (from 1 to 2) t^2 dt \approx 7/3 = 2.3333[/tex]

So the left-sum approximation is still an underestimate, but it is closer to the true value than the previous approximation.

The right-sum approximation is still an overestimate, but it is also closer to the true value than the previous approximation.

For similar question on rectangles.

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Related Questions

a ball that is dropped from a window hits the ground in 7 seconds. how high is the window? (give your answer in feet; note that the acceleration due to gravity is 32 ft/s.)

Answers

The ball was dropped from a window that is 784 feet high. To determine the height of the window from which the ball was dropped, we can use the formula for free fall: h = 0.5 * g * t²


The formula for free fall is :  h = 0.5 * g * t² ,

where h is the height, g is the acceleration due to gravity (32 ft/s²), and t is the time it takes to hit the ground (7 seconds).

Given below the steps to calculate how high the window is :

Plug in the values to the equation:
h = 0.5 * 32 * (7²)Calculate the square of the time
7² = 49Multiply the values
h = 0.5 * 32 * 49Calculate the height
h = 16 * 49
h = 784 feet

So, the ball was dropped from a window that is 784 feet high.

To learn more about  dropped : https://brainly.com/question/24746268

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