Can someone explain b) to me. I don't quite fully understand why you take out a factor of 1000 . Why wouldn't you rearrange 7100 to = 8-9x and solve for x and sub that back in?
BTW theres an error in the MS for that Question. Its supposed to be 10³√7.1 not 10³√71. Try typing on the calc. Now to explain i "think" x should always be less than 1. Someone correct me.
Dont know why but if subbing 7100 gives you the wrong answer compared to when you type ³√7100 on calculator try turing it to surds. ³√7100 = ³√7.1 × ³√1000 = 10³√7.1 Equate 10³√(9-8x) = 10³√7.1 9-8x = 7.1 x = 0.1
Really confused. trying to do this integral: 1/(3x+3). Pretty sure the answer is (1/3)*ln(3x+3) but surely if you take out 1/3 you get 1/(x+1) and then as the top is the differetial of the bottom it would be (1/3)*ln(x+1). Any help appreciated. Good luck everyone!
ln(3x + 3) = ln3 + ln(x + 1) All you're doing with the first case is adding a specific constant bit
For integration by parts to work in this question would you need to expand out the brackets first then solve them as two separate integrals? Like this:
∫04ππx2dx+∫04ππx2(sin2x)dx
I tried to to this question last night without expanding first and got most of the answer correct but the first term was missing, guessing this is why...
For integration by parts to work in this question would you need to expand out the brackets first then solve them as two separate integrals? Like this:
∫04ππx2dx+∫04ππx2(sin2x)dx
I tried to to this question last night without expanding first and got most of the answer correct but the first term was missing, guessing this is why...
You would need to integrate by parts twice for that wouldn't you?
but what have i done wrong? one of the 2 ways is wrong as they dont equal each other.
It's not wrong, it's just a specific case. Normally you would put a constant of integration, right? If you take out the constant in the 3x + 3 version, you can then amalgamate that with some arbitrary constant to get another arbitrary constant; i.e. if you write the integral as
(1/3)ln(3x + 3) + c this is 1/3(ln(3) + ln(x + 1)) + c = (1/3)ln3 + (1/3)ln(x + 1) + c
Now this can be written as (1/3)ln(x + 1) + k because the c and the (1/3)ln3 are constants
If you put the answer (1/3)ln(x + 1) this is more correct to be honest (although having a plus c is always the most correct)
Really confused. trying to do this integral: 1/(3x+3). Pretty sure the answer is (1/3)*ln(3x+3) but surely if you take out 1/3 you get 1/(x+1) and then as the top is the differetial of the bottom it would be (1/3)*ln(x+1). Any help appreciated. Good luck everyone!
Both are equivalent because the 1/3ln(3x+3) is 1/3(ln3+ln(x+1)) When you put your limits in you have ln3 and -ln3 so the answer is the same for both
BTW theres an error in the MS for that Question. Its supposed to be 10³√7.1 not 10³√71. Try typing on the calc. Now to explain i "think" x should always be less than 1. Someone correct me.
Dont know why but if subbing 7100 gives you the wrong answer compared to when you type ³√7100 on calculator try turing it to surds. ³√7100 = ³√7.1 × ³√1000 = 10³√7.1 Equate 10³√(9-8x) = 10³√7.1 9-8x = 7.1 x = 0.1
I'm beginning to grasp why this works now. How Edexcel thinks about putting questions like these in, hurt me.
BTW theres an error in the MS for that Question. Its supposed to be 10³√7.1 not 10³√71. Try typing on the calc. Now to explain i "think" x should always be less than 1. Someone correct me.
Dont know why but if subbing 7100 gives you the wrong answer compared to when you type ³√7100 on calculator try turing it to surds. ³√7100 = ³√7.1 × ³√1000 = 10³√7.1 Equate 10³√(9-8x) = 10³√7.1 9-8x = 7.1 x = 0.1
When a gradient is parallel to the x axis and you have eg dy/dx = (3x+ y)/(2y+1) is the 3x + y = 0 or then 2y+1? thanks
If the gradient is equal to 0, the top line (numerator) is equated to 0. If a line is parallel to the y-axis then the gradient is undefined, and you only get an undefined value when 0 is on the bottom of a fraction, therefore you equate 2y+1= 0 (denominator)
It's not wrong, it's just a specific case. Normally you would put a constant of integration, right? If you take out the constant in the 3x + 3 version, you can then amalgamate that with some arbitrary constant to get another arbitrary constant; i.e. if you write the integral as
(1/3)ln(3x + 3) + c this is 1/3(ln(3) + ln(x + 1)) + c = (1/3)ln3 + (1/3)ln(x + 1) + c
Now this can be written as (1/3)ln(x + 1) + k because the c and the (1/3)ln3 are constants
If you put the answer (1/3)ln(x + 1) this is more correct to be honest (although having a plus c is always the most correct)