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Mathematics: Post your doubts here!

Binyamine

Binyamine

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rockerzregmi said:
guys can you solve this question ?
integration of (xe^2x) dx
Click to expand...

We will have to solve by part. Consider Late, x is algebraic while e ^ ( 2*x) is exponential.

Hence; u = x ; v = ( e ^ ( 2* x ) ) / 2
du/dx = 1 ; dv/dx = e ^ ( 2*x)

integration of (xe^2x) dx = u * v - integration of v du
= x * ( e ^ ( 2* x ) ) / 2 - integration of 1 * ( e ^ ( 2* x ) ) / 2
= x * ( e ^ ( 2* x ) ) / 2 - ( e ^ ( 2* x ) ) / 4
 
sweetiepie

sweetiepie

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Q. In How Many Years a sum of Rs 3000 would amount Rs 6130.43 at 6% compounded quarterly ?
Q. Find the amount of Rs 250 invested at the end of each of 5 successive years at 6% interest compounded annually ?
plz help me as soon as possible with all steps :(
 
salvatore

salvatore

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Binyamine said:
LOL, thanks for saying that i am awesome. hahaha but i want to become LEGEND...wait for it....LEGENDAAAAARYYYYYYYYYY. [ Dialogue of Barney Stinson ]

Well to answer 11 (iii), i presume that you have successfully tackled part (i) and part (ii), i.e. you already know the coordinates of A, B and the equation of the normal to the curve at C.

Now, the first step is to find coordinate of C, by solving simultaneously the equation of curve and the equation of the curve at C. Just find its x coordinate. Infact you will get two values of x, reject the value of x which is smaller than the x coordinate of point A.

Let us say that Area H = integration of the equation of normal to the curve with lower limit the x ordinate of coordinate C and upper limit the x ordinate of coordinate B

Let us say that Area K = integration of the equation of the curve with lower limit the x ordinate of coordinate C and upper limit the x ordinate of coordinate B

Try to visualise what the above area represents and you would come to the conclusion that

the area of the shaded region = Area H - Area K

Hope that it helps, if it does then make du'a for me and for all the oppressed muslims wherever they are especially the Syrian people and Afghan people.
Click to expand...
Lool.. dw, u'll become a legend ;)
The part about Area K is correct.. but I think you have made an error in Area H.
According to the marking scheme, you are supposed to find the area of a trapezium and subtract it from area K. Now where did the trapezium come from?? :confused: :(

And yeah.. may god help the oppressed Muslims around the world!
 
Binyamine

Binyamine

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salvatore said:
Lool.. dw, u'll become a legend ;)
The part about Area K is correct.. but I think you have made an error in Area H.
According to the marking scheme, you are supposed to find the area of a trapezium and subtract it from area K. Now where did the trapezium come from?? :confused: :(

And yeah.. may god help the oppressed Muslims around the world!
Click to expand...

My friend, analyse well, put your hand on the line which is normal to the curve at C and go down from the value of x of point A and value of X for point C. You see the area H that i talked of is infact that area of trapezium that is mentioned in the marking scheme.
 
salvatore

salvatore

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Binyamine said:
My friend, analyse well, put your hand on the line which is normal to the curve at C and go down from the value of x of point A and value of X for point C. You see the area H that i talked of is infact that area of trapezium that is mentioned in the marking scheme.
Click to expand...
Oh yeah.. hadn't figured that out earlier.. that was dumb of me.
Thanks for your help mate!
 
salvatore

salvatore

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VampGirl said:
Plz help me with q:8 (ii) of oct/nov/05 p1

Thnx
Click to expand...
Inverse:

(i) Make equate the function to y
-----> y = (2x − 3)^3 − 8

(ii) Make x the subject
-----> y + 8 = (2x - 3)^3
(y + 8)^1/3 = (2x - 3)^1/3.... ^1/3 refers to cubic root
(y + 8)^1/3 = 2x - 3
(y + 8)^1/3 + 3 = 2x
x = ( (y + 8)^1/3 + 3 ) / 2

(iii) The switch the y with x.. this becomes the inverse
---> f−1(x) = ( (x + 8)^1/3 + 3 ) / 2

Domain:

The range of the function becomes the domain of the inverse.. and vice versa
Hence, use the domain of the function to find the range i.e 2 ≤ x ≤ 4. Substitute the values in the expression:

Lower limit:
y = (2 (2) − 3)3 − 8
= -7

Upper limit:
y = (2 (4) − 3)3 − 8
=117

The domain becomes -7 ≤ x ≤ 117 :)​
 
VampGirl

VampGirl

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salvatore said:
Inverse:

(i) Make equate the function to y
-----> y = (2x − 3)^3 − 8

(ii) Make x the subject
-----> y + 8 = (2x - 3)^3
(y + 8)^1/3 = (2x - 3)^1/3.... ^1/3 refers to cubic root
(y + 8)^1/3 = 2x - 3
(y + 8)^1/3 + 3 = 2x
x = ( (y + 8)^1/3 + 3 ) / 2

(iii) The switch the y with x.. this becomes the inverse
---> f−1(x) = ( (x + 8)^1/3 + 3 ) / 2

Domain:

The range of the function becomes the domain of the inverse.. and vice versa
Hence, use the domain of the function to find the range i.e 2 ≤ x ≤ 4. Substitute the values in the expression:

Lower limit:
y = (2 (2) − 3)3 − 8
= -7

Upper limit:
y = (2 (4) − 3)3 − 8
=117

The domain becomes -7 ≤ x ≤ 117 :)​
Click to expand...
Thank u :)
 
K

kronix6

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Need help in this question plz.

given that sinQ - cosQ /sinQ + cosQ= 6sinQ/cosQ


Find the exact values of tanQ
 
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