Join TSR now and get all your revision questions answeredSign up now
    • Thread Starter
    Offline

    2
    ReputationRep:
    What do I have to do for this question? :/

    Attached Images
     
    • Study Helper
    Offline

    18
    ReputationRep:
    (Original post by creativebuzz)
    What do I have to do for this question? :/

    I'm guessing you've done part A and you've got a problem with part B. So obviously, there's no one-step method to get the shaded area. However, if I tell you that you can get that area by subtracting the area under one line from the area under another, does that help?
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Chlorophile)
    I'm guessing you've done part A and you've got a problem with part B. So obviously, there's no one-step method to get the shaded area. However, if I tell you that you can get that area by subtracting the area under one line from the area under another, does that help?

    For part (a) I differentiated to get

    dy/dx = 5 - 4x

    I then subbed in the 1

    to get a gradient of dy/dx = 1

    then I did the negative reciprocal to get the normal gradient, which means my gradient is -1

    I then subbed that into the line equation to get

    y - 3 = -(x-1)

    y-3=-x+1

    x+y-4=0

    Is that correct or have I gone wrong somewhere?

    As for (b) I thought of doing the area of trapezium minus the area under the line and the curve! Is that correct or?
    • Study Helper
    Offline

    18
    ReputationRep:
    (Original post by creativebuzz)
    For part (a) I differentiated to get

    dy/dx = 5 - 4x

    I then subbed in the 1

    to get a gradient of dy/dx = 1

    then I did the negative reciprocal to get the normal gradient, which means my gradient is -1

    I then subbed that into the line equation to get

    y - 3 = -(x-1)

    y-3=-x+1

    x+y-4=0

    Is that correct or have I gone wrong somewhere?

    As for (b) I thought of doing the area of trapezium minus the area under the line and the curve! Is that correct or?
    For (a) that is correct, well done. For (b) yes that's basically right, the shaded area is the area under the line (i.e. a trapezium) minus the area under the curve. Make sure the limits are right though!
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by Chlorophile)
    For (a) that is correct, well done. For (b) yes that's basically right, the shaded area is the area under the line (i.e. a trapezium) minus the area under the curve. Make sure the limits are right though!
    Awesome! I've found the area of the trapezium


    As for the limits, I set the equation of the curve equal to zero and my result was x = 0 and x=5/2! Is that correct?
    Attached Images
      
 
 
 
Poll
If you won £30,000, which of these would you spend it on?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Quick reply
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.