Have to show area is 24, I get 12?? [Integration Issue] Watch

OJ Emporium
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#1
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The question: https://gyazo.com/63d9629626e647a26e5bc5c9ce77bc67
https://gyazo.com/88d2200d199bce319314ddc050d4b80b

What I did:

Found the derivative to in order to find the gradient which was 6
y-15 = 6x-24
y = 6x-9

After rearranging when each point(intercepts) is 0, x = 3/2 and y = -9

Integrated the area of the curve to get 36
Then from that I tried to find the area of the triangle underneath the graph by doing bxh/2:
base is 4
height is 24 (inbetween -9 and 15)
(4x24)/2 is 48 and did 36-48 to get -12 which was just blatantly wrong
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mqb2766
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A bit hard to read the question, but remember that area below y=0 is negative. This would affect both the curve and the choice of triangle.
(Original post by OJ Emporium)
The question: https://gyazo.com/63d9629626e647a26e5bc5c9ce77bc67
https://gyazo.com/88d2200d199bce319314ddc050d4b80b

What I did:

Found the derivative to in order to find the gradient which was 6
y-15 = 6x-24
y = 6x-9

After rearranging when each point(intercepts) is 0, x = 3/2 and y = -9

Integrated the area of the curve to get 36
Then from that I tried to find the area of the triangle underneath the graph by doing bxh/2:
base is 4
height is 24 (inbetween -9 and 15)
(4x24)/2 is 48 and did 36-48 to get -12 which was just blatantly wrong
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OJ Emporium
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(Original post by mqb2766)
A bit hard to read the question, but remember that area below y=0 is negative. This would affect both the curve and the choice of triangle.
Here: https://media.discordapp.net/attachm...400&height=208 Apologies for the low quality-ness of the other picture lol.

But for the area I realised I could simply do 15-9 to get 6, multiply that by 4 and halve it to get 12
do 36-12 to get 24 but i think this is illegal in maths terms.
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STEM.lover
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Is that 5x^3/2?
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mqb2766
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If that's the first term, then yes I couldn't read it. Why not shift your graph up by the amount such that the tangent line passes through the origin, then subtract the triangle?
Post you working if it's not right.
(Original post by STEM.lover)
Is that 5x^3/2?
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STEM.lover
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Hey I’ve found out the correct way of doing this

Everything you did first was right you have the line of equation y=6x-9

When x= 0 y=-9
When y= 0 x = 1.5

Integrating finds the area between the curve at the x axis therefore the little triangle under the curve has to be calculated separately and added on to the final answer. You do this by doing 9*1.5 / 2, the intersection of x and y axis I stated before. This is 6.75

Then you need to integrate between the limits of 4 and 0 you should get an answer of 36.

Then you take away the triangle under the straight line calculated by 4-1.5 = 2.5
2.5 * 15/ 2 which is 18.75

6.75 + (36-18.75) = 24

Hope this helps let me know if you don’t understand anything
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OJ Emporium
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(Original post by STEM.lover)
Hey I’ve found out the correct way of doing this

Everything you did first was right you have the line of equation y=6x-9

When x= 0 y=-9
When y= 0 x = 1.5

Integrating finds the area between the curve at the x axis therefore the little triangle under the curve has to be calculated separately and added on to the final answer. You do this by doing 9*1.5 / 2, the intersection of x and y axis I stated before. This is 6.75

Then you need to integrate between the limits of 4 and 0 you should get an answer of 36.

Then you take away the triangle under the straight line calculated by 4-1.5 = 2.5
2.5 * 15/ 2 which is 18.75

6.75 + (36-18.75) = 24

Hope this helps let me know if you don’t understand anything
Yikes so the way I did it was purely coincidental that I got 24

But thank you, I did something similar to this but got some crazy answer
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STEM.lover
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(Original post by OJ Emporium)
Yikes so the way I did it was purely coincidental that I got 24

But thank you, I did something similar to this but got some crazy answer
Yeah I saw that, you combined both of your triangles instead of splitting them up. The most important thing to remember is integrating is the area bound between curve and x axis. If it splits the axis two separate integrals are needed.
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