user3.14159
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Help with this question please. There's a diagram too, any help would be appreciatedI'm really struggling.

the region enclosed between the curves y=e^x y=6-e^x/2 and the line x=0 shown shaded in the diagram below sow that the exact area of the shaded region is 6ln4-5 fully justify your answer
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mqb2766
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(Original post by user3.14159)
Help with this question please. There's a diagram too, any help would be appreciatedI'm really struggling.

the region enclosed between the curves y=e^x y=6-e^x/2 and the line x=0 shown shaded in the diagram below sow that the exact area of the shaded region is 6ln4-5 fully justify your answer
What are you stuck with (and the diagram would help ...)? The integration, finding the limits, ..
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Muttley79
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(Original post by user3.14159)
Help with this question please. There's a diagram too, any help would be appreciatedI'm really struggling.

the region enclosed between the curves y=e^x y=6-e^x/2 and the line x=0 shown shaded in the diagram below sow that the exact area of the shaded region is 6ln4-5 fully justify your answer
Please post how far you've got.
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user3.14159
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(Original post by mqb2766)
What are you stuck with (and the diagram would help ...)? The integration, finding the limits, ..
Hi, thanks for replying. I'm having trouble with where to begin - exponentials is something I've struggled with for a while. I'll try to upload a photo (I'm like an 80 year old when it comes to technology). It's from the aqa released documents on integration.
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laurawatt
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(Original post by user3.14159)
Hi, thanks for replying. I'm having trouble with where to begin - exponentials is something I've struggled with for a while. I'll try to upload a photo (I'm like an 80 year old when it comes to technology). It's from the aqa released documents on integration.
Got it here
Name:  9673381F-B2E9-4293-936E-D9AFB9B30A4A.jpeg
Views: 12
Size:  33.5 KB
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mqb2766
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(Original post by user3.14159)
Hi, thanks for replying. I'm having trouble with where to begin - exponentials is something I've struggled with for a while. I'll try to upload a photo (I'm like an 80 year old when it comes to technology). It's from the aqa released documents on integration.
You obviously need to know how to integrate exponentials to answer the question. Similarly, if R is defined by where they interect, you'll need to find that point. Post what youve tried.
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user3.14159
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(Original post by laurawatt)
Got it here
Name:  9673381F-B2E9-4293-936E-D9AFB9B30A4A.jpeg
Views: 12
Size:  33.5 KB
Yes that's it. You're a life saver I've been trying to figure out how to attach a photo forever!!
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laurawatt
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Does this look familiar?
Name:  80A3A0AB-6A74-4296-A675-ED3D51650C80.jpeg
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It would be the easiest way to solve it
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user3.14159
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(Original post by mqb2766)
You obviously need to know how to integrate exponentials to answer the question. Similarly, if R is defined by where they interect, you'll need to find that point. Post what youve tried.
Would I just integrate the values or use substitution?
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mqb2766
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(Original post by user3.14159)
Would I just integrate the values or use substitution?
Can you integrate the functions individually?
As laurawatt says, if you're ok with that, you can consider R as being the difference of two areas.
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user3.14159
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(Original post by mqb2766)
Can you integrate the functions individually?
As laurawatt says, if you're ok with that, you can consider R as being the difference of two areas.
I think I may have done it wrong but is it just 6x-2e^x/2 + C
Last edited by user3.14159; 1 month ago
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mqb2766
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(Original post by user3.14159)
I think I may have done it wrong but is it just 6x-2e^x/2 + C
You can always differentiate your answer to check, but that looks ok for the indefinite integral of the "harder" function.
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user3.14159
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(Original post by mqb2766)
You can always differentiate your answer to check, but that looks ok for the indefinite integral of the "harder" function.
Yeah, I'm alright with that integration but I'm not sure what the next steps are
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mqb2766
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(Original post by user3.14159)
Yeah, I'm alright with that integration but I'm not sure what the next steps are
Tbh, finding the area of a region like this is a reasonably common question and if you're attempting exam practice questions, you should have come across similar ones before? Laurawatt's post gave the formula, so it involves integrating the difference of two functions between a and b. What does that mean for this question - what is the upper / lower function, what are the limits of the definite integral?
Last edited by mqb2766; 1 month ago
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user3.14159
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(Original post by mqb2766)
Tbh, finding the area of a region like this is a reasonably common question and if you're attempting exam practice questions, you should have come across similar ones before? Laurawatt's post gave the formula, so it involves integrating the difference of two functions between a and b. What does that mean for this question - what is the upper / lower function, what are the limits of the definite integral?
I'm not sure how to determine the upper and lower, limits of the definite integral, the question is posted on this thread but I don't know how to find these out?
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MiladA
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(Original post by user3.14159)
Yeah, I'm alright with that integration but I'm not sure what the next steps are
You need the limits of integration. Find the x coordinate point of intersection of the two curves and then set your limits of integration to 0 (lower bound) and the x coordinate (upper bound). There is no need for + C as it is a definite integral.

The integration which you did above is correct but you also need to take into account the fact that the region is bounded by two curves not just one. So you need to do the integral of (f(x) - g(x)) dx which you can easily visualise which is f(x) and which is g(x) given the diagram and the area (region) required.
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mqb2766
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(Original post by user3.14159)
I'm not sure how to determine the upper and lower, limits of the definite integral, the question is posted on this thread but I don't know how to find these out?
Do you understand what they represent? The x-values where the integral starts and finishes. One is "obvious" as its simple and given in the question. The other needs to be calculated using a fact.
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user3.14159
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(Original post by mqb2766)
Do you understand what they represent? The x-values where the integral starts and finishes. One is "obvious" as its simple and given in the question. The other needs to be calculated using a fact.
One of them would be x=0 I think
Last edited by user3.14159; 1 month ago
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mqb2766
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(Original post by user3.14159)
One of them would be x=0 I think
of course. What gives the upper limit?
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user3.14159
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(Original post by mqb2766)
of course. What gives the upper limit?
Is it the intercept?
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