Improper rational function is defined as rational function with highest degree of x in the numerator greater than or equal to to the highest degree of x in the denominator. Polynomial (long) division is needed before the integration of the improper rational function
The formula : integration of 1/(x^2+a^2) w.r.t x =(1/a)arc tan(x/a) is given in the MF19 formulae list, hence differentiation of arc tan(x/a)=a/(x^2+a^2). I believed that my method will be acceptable as it is in the MF19. Fyi, the method in mark scheme is differentiation of implicit equation
For the Integration by parts, u is given according to this rule LIATE, L:ln function, I: inverse trigo, A:algebraic(polynomial), T:trigonometric function, E: exponential function
@@lakshagungabissoon8585 Yes, we integrate x. Integration by part is applied here. u= arc tan 2x and du/dx=2/(1+4x^2), dv/dx=x, v is integration of dv/dx, hence v=(x^2)/2
OMG I just finished 1 hour ago but I was so confused with the ms. Thx for updating❤❤❤
You're most welcome!
Thank you very much
You are welcome! Please continue supporting this channel by sharing it with everyone. Thank you
Ohhhh that’s what shortest distance means
Shortest distance means perpendicular distance
is it a must we show working for quadratic formula
Better to show
1:25:06 why did you use long division here? why can't we directly integrate it?
Improper rational function is defined as rational function with highest degree of x in the numerator greater than or equal to to the highest degree of x in the denominator. Polynomial (long) division is needed before the integration of the improper rational function
Miss, for 10(a) the mark scheme solves it by using sec^2x. would you still get the mark doing your method?
The formula : integration of 1/(x^2+a^2) w.r.t x =(1/a)arc tan(x/a) is given in the MF19 formulae list, hence differentiation of arc tan(x/a)=a/(x^2+a^2). I believed that my method will be acceptable as it is in the MF19. Fyi, the method in mark scheme is differentiation of implicit equation
thank you for this video miss, for number 6 part b, how would we have solved it if we both our F(x) was positive or both was negative?
When you are asked to verify the root lies between a and b, then we need to apply the sign change rule, which is f(a)>0 and f(b)
queen
Thanks for your support. 🙏
isnt 5pi/6 out of the range for arg(v/u) ?
-pi
Miss, for qs 10b), how come u= tan inverse 2x and v= x ? Isn't it we differentiate the the x ?? 😢
For the Integration by parts, u is given according to this rule LIATE, L:ln function, I: inverse trigo, A:algebraic(polynomial), T:trigonometric function, E: exponential function
@MathWorld-yp9od shouldn't you integrate x first
@@lakshagungabissoon8585 Yes, we integrate x. Integration by part is applied here. u= arc tan 2x and du/dx=2/(1+4x^2), dv/dx=x, v is integration of dv/dx, hence v=(x^2)/2
Following ❤
Thank you.
What would be the maximum value in 7 b
Take the least value in 7b and add with the diameter of circle to get the maximum value
May you please do 32 may june 9709?
Sure, stay tuned for more videos. Thank you
Nearly spotted a mistake in qs 9 for vector hahaha
😘😘