You must show tlhe detailed steps for all questions. No credit will be given if you do not show your work(a) Use the Itnit process to find the area of the region bounded by the graph of the function and the r-axis over the giveninterval; (b) use the fundamental theorem of caleulus to verify your result; (c) find the average value of the function overthe given interval2. f(z) 2r3 +2 +3, [0,3

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Asked Nov 22, 2019
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can someone please show me how to solve this problem. I don't know if I'm doing it right. I got 117/2 for the area in part a but i don't know how to solve part b or c. 

You must show tlhe detailed steps for all questions. No credit will be given if you do not show your work
(a) Use the Itnit process to find the area of the region bounded by the graph of the function and the r-axis over the given
interval; (b) use the fundamental theorem of caleulus to verify your result; (c) find the average value of the function over
the given interval
2. f(z) 2r3 +2 +3, [0,3
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You must show tlhe detailed steps for all questions. No credit will be given if you do not show your work (a) Use the Itnit process to find the area of the region bounded by the graph of the function and the r-axis over the given interval; (b) use the fundamental theorem of caleulus to verify your result; (c) find the average value of the function over the given interval 2. f(z) 2r3 +2 +3, [0,3

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Step 1

(b)

Given,

f(x) = 2x3 x2 +3 ,
хе [0,3]
We know that
The area of the region bounded the curve y = f(x) and x - axis on
the interval [a, b] is given by,
-f(x) dx
a
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f(x) = 2x3 x2 +3 , хе [0,3] We know that The area of the region bounded the curve y = f(x) and x - axis on the interval [a, b] is given by, -f(x) dx a

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Step 2

Thus,

...
substituting f(x) = 2x3 x2 3, a = 0 and b = 3 in the above
formula we get,
(2x3x23 dx
A
3
(x2) dx
(2x3) dx |
(3) dx
0
for constant k
3
x2 dx3
x3 dx
= 2
dx
kf(x)dx
<f(x)dx
=
31
xn+1
3x1
for n -1
= 2
4
0
H
х^ dx
n + 1
0
34
33
3 3
2
117
58.5 square units
2
units
Therefore, the area bounded by f(x) is 58.5 sq
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substituting f(x) = 2x3 x2 3, a = 0 and b = 3 in the above formula we get, (2x3x23 dx A 3 (x2) dx (2x3) dx | (3) dx 0 for constant k 3 x2 dx3 x3 dx = 2 dx kf(x)dx <f(x)dx = 31 xn+1 3x1 for n -1 = 2 4 0 H х^ dx n + 1 0 34 33 3 3 2 117 58.5 square units 2 units Therefore, the area bounded by f(x) is 58.5 sq

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Math

Calculus