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dreamingtruth dreamingtruth
wrote...
Posts: 77
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10 years ago
1.Sketch the region bounded by the graphs of the functions and find the area of the region. f(x)= 3-2x-x^2, g(x)=x+3?

2. Find the area bounded by the parabola x=8+2y-y^2 the y axis and the lines y=-1 and y=3?

3. Sketch the region bounded by the graphs of the functions and find the area of the region?
f(x)= x^3-2x+1, g(x)= -2x, x=1

4. Sketch the region bounded by the graphs of the functions and find the area of the region f(y)= y(2-y) g(y)=-y?

As you can see, these questions are all based around the same topic. Help with any of them would be amazing!

And here is the trapezoidal question:

5.f(x) is continuous on the closed interval [1, 5] and goes through the following points: Use the trapezoidal rule, with n = 4 to approximate the area under the curve from x = 1 to x = 5.
 
x---------- 1 ------------ 2 ----------- 3 ------------ 4 ------------- 5
f(x) ------ 1.62 --------4.15 ------- 7.50 -------- 9.0 ---------- 12.13


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wrote...
Educator
10 years ago
1.Sketch the region bounded by the graphs of the functions and find the area of the region. f(x)= 3-2x-x^2, g(x)=x+3?

1) Find out where the two functions intersect. This can be done by making the two equations equation to one another.

f(x) = g(x)

2) You have to find the integral of the first function and the integral of the second function, and then subtract the two integrals.

f'(x)=−x^3/3 − x^2 + 3x
g'(x)= x^2/2 + 3x

3) To find the region bounded by the two functions, find the integral between the two intersection points (x1 will be your lower bound, and x2 will be your upper bound).

\({\int_{a}^b}\)

Quote
2. Find the area bounded by the parabola x=8+2y-y^2 the y axis and the lines y=-1 and y=3?

Y-axis means that you have a function x = 0 (wouldn't call it a function, but more so a vertical line).

\({\int_{-1}^3}\)[8y+y^2-y^(3)/3]

= 8*4 + (9 - 1) - (27 + 1)/3
= 32 + 8 - 28/3
= 30.666667

Please, two questions per thread... That's all I can do for now.
wrote...
10 years ago
Thank you for your help! I understood what you did perfectly (:
wrote...
Educator
10 years ago
Thank you for your help! I understood what you did perfectly (:

You're welcome.
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