The net signed area between the curve of the function f(x) = x - 1 and the x-axis over the interval [-7, 3] is -41.
To find the net signed area between the curve of the function f(x) = x - 1 and the x-axis over the interval [-7, 3], we need to integrate the function from -7 to 3 and take into account the signed area.
The integral of f(x) = x - 1 over the interval [-7, 3] is given by:
∫[-7, 3] (x - 1) dx
Evaluating this integral, we get:
[tex]∫[-7, 3] (x - 1) dx = [1/2 * x^2 - x] [-7, 3]\\= [(1/2 * 3^2 - 3) - (1/2 * (-7)^2 - (-7))][/tex]
= [(9/2 - 3) - (49/2 + 7)]
= [9/2 - 3 - 49/2 - 7]
= (-27/2) - (55/2)
= -82/2
= -41
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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.)
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 :
So, the ball was dropped from a window that is 784 feet high.
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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.
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.
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