8) Find the area of the region contained by the graphs of 8- y =.y = 8x, and y = x, shown in the diagram at right: (1,8) Solution? -6 Comparing gradients, we can tell that the red line is y = 8x and the blue line is y = x. The black curve is y = 4- To find the area between the curves we must integrate, and also must subtract (top function) – (bottom function). On the interval ]0,1[ the top function is y = 8x, but on the interval ]1,2[ the top function is y = So we will find two 2 (2,2) definite integrals and add them together to find the total (0. 0) area. 05 1.5 8х — х dx + 8 x dx (1,8) 1 8 dx + 8x-2 — х dx = [8x]5 + 1-X8- 1 - 10 ()-(-) 22 8 [8] + 8 - (22) 2 (0, 0) = 8 + (-6) – (-8.5) = 10.5 units2 05 1.5
8) Find the area of the region contained by the graphs of 8- y =.y = 8x, and y = x, shown in the diagram at right: (1,8) Solution? -6 Comparing gradients, we can tell that the red line is y = 8x and the blue line is y = x. The black curve is y = 4- To find the area between the curves we must integrate, and also must subtract (top function) – (bottom function). On the interval ]0,1[ the top function is y = 8x, but on the interval ]1,2[ the top function is y = So we will find two 2 (2,2) definite integrals and add them together to find the total (0. 0) area. 05 1.5 8х — х dx + 8 x dx (1,8) 1 8 dx + 8x-2 — х dx = [8x]5 + 1-X8- 1 - 10 ()-(-) 22 8 [8] + 8 - (22) 2 (0, 0) = 8 + (-6) – (-8.5) = 10.5 units2 05 1.5
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
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Chapter1: Functions And Models
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Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Binomial is an algebraic expression of the sum or the difference of two terms. Before knowing about binomial distribution, we must know about the binomial theorem.
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Identify if there are any errors in the solutions, and state what they are. Thanks
![8) Find the area of the region contained by the graphs of
8-
y =y = 8x, and y = x, shown in the diagram at right:
(1,8)
Solution?
-6
Comparing gradients, we can tell that the red line is y = 8x
and the blue line is y = x. The black curve is y =
4-
To find the area between the curves we must integrate, and
also must subtract (top function) – (bottom function).
On the interval ]0,1[ the top function is y = 8x, but on the
interval ]1,2[ the top function is y = So we will find two
2
(2,2)
definite integrals and add them together to find the total
(0. 0)
area.
05
1.5
8х — х dx +
8
x dx
(1,8)
1
8 dx +
8x-2 — х dx
= [8x]5 +
1-X8-
1
- 10 ()-(-)
22
8
[8] +
8 -
(22)
2
(0, 0)
= 8 + (-6) – (-8.5) = 10.5 units2
05
1.5](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8ce2e611-8c07-4409-a463-d0a61141ee83%2F42912acf-fd81-4e61-8b91-0720ceaf1330%2F0r7dkej_processed.png&w=3840&q=75)
Transcribed Image Text:8) Find the area of the region contained by the graphs of
8-
y =y = 8x, and y = x, shown in the diagram at right:
(1,8)
Solution?
-6
Comparing gradients, we can tell that the red line is y = 8x
and the blue line is y = x. The black curve is y =
4-
To find the area between the curves we must integrate, and
also must subtract (top function) – (bottom function).
On the interval ]0,1[ the top function is y = 8x, but on the
interval ]1,2[ the top function is y = So we will find two
2
(2,2)
definite integrals and add them together to find the total
(0. 0)
area.
05
1.5
8х — х dx +
8
x dx
(1,8)
1
8 dx +
8x-2 — х dx
= [8x]5 +
1-X8-
1
- 10 ()-(-)
22
8
[8] +
8 -
(22)
2
(0, 0)
= 8 + (-6) – (-8.5) = 10.5 units2
05
1.5
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