for q9, the requirement is to find the area shaded – you can solve this in relation with the x axis or y axis and you integrate not differentiate.. the most common solution is with the x-axis..
for solution with x axis –> area = integral from 0 to 3 ( y curve – y tangent)dx. in general area=integral from a to b of y dx and is representing the area between the curve/line, the vertical lines x=a and x=b and the x axis.. so in our case you have the area shaded= area under the curve – area under tangent with the same limits (the solution is shown in my work http://justpastpapers.com/wp-content/uploads/2014/05/IMG_3477.jpg)
for solution with y axis –> you need to find the y values for both the tangent and curve when x=0 and to make x the subject of the formula in both, so area = integral from 11/2 to 7 (x tangent) dy- integral from 6 to 7 (x curve) dy.. in general area=integral from c to d of x dy and is representing the area between the curve/line, the horizontal lines y=c and y=d and the y axis
the second method takes more time so the first one is better to be used in this case.. both give the same answer
while opening the page for question (3), the page for question (2), is actually opening.
solved
Sir Qs 3 and 5 are not available could you kindly send them
you can access them now
click on question 3 showing question 2 :/
now is working
is it possible that you do the workings for 2013 may/june papaer 12 and oct/nov…we havea paper1 exam tomorrow..
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Thank u so much! U helped me greatly☺
In question 9, isnt the area due to y-axis, can you please explain me how to differentiate if it is due to x-axis or y-axis?
for q9, the requirement is to find the area shaded – you can solve this in relation with the x axis or y axis and you integrate not differentiate.. the most common solution is with the x-axis..
for solution with x axis –> area = integral from 0 to 3 ( y curve – y tangent)dx. in general area=integral from a to b of y dx and is representing the area between the curve/line, the vertical lines x=a and x=b and the x axis.. so in our case you have the area shaded= area under the curve – area under tangent with the same limits (the solution is shown in my work http://justpastpapers.com/wp-content/uploads/2014/05/IMG_3477.jpg)
for solution with y axis –> you need to find the y values for both the tangent and curve when x=0 and to make x the subject of the formula in both, so area = integral from 11/2 to 7 (x tangent) dy- integral from 6 to 7 (x curve) dy.. in general area=integral from c to d of x dy and is representing the area between the curve/line, the horizontal lines y=c and y=d and the y axis
the second method takes more time so the first one is better to be used in this case.. both give the same answer