Before we get into the questions, anybody know how to integrate a polynomial???? I tink we have to add 1 to the power and divide by the power. for example find the integral of 2x^5. If we add 1 to the power we wud get 6, and divide by the power which is 6, so we wud get 2/6x^6 + c. ENT????
well for this question: i think that yuh hav to integrate to find the area under the curve after the time = 4. therefore: the integral answer wud be: 2t^3+ 2t^2 +(t^2/2) + c and plug in t=4 = 128 + 32 + 8 = 168units
A particle moves along the x-axis with acceleration a = 2t - 3 m/s^2. At time t = 0, it is at the origin and moving with a speed of 4 m/s in the positive direction. Find formulas for its velocity v and position s, and determine where it changes direction and where it is moving to the left.
first integration is the total opposite of defrentiation, like (+) and (-). Hence when u defrentiate y=4x^2 and u get :dy/dx = 8x, and now if you integrate;dy/dx = 8x you will get :y=4x^2.
MISS!! with respect ot your question, we have to keep in mind v = ds/dt and a = dv/dt. Therefore dv/dt = 2t - 3 m/s^2 so to get v we must integrate dv/dt. we wud get v = t^2 - 3t + c. EVERYBODY KNOW HOW TO INTEGRATE RIGHT..SHOCKS I HAVE TO GO AND MEET A GIRL NOW. IL GET BACK TO U ALL LATER... i
hey hellion i think i wud give your quest. a try but correct me if i am wrong well first of all i think you will have to substitute where t is put for and solve.
so it will be
6t^2 + 4t + t. where t=4 6(4)^2 + 4(4)+ 4
remeber your BODMAS rule befor you go any further solve brackets first so it will be 4*4whichwas really the 4 squared and this equals 16 then multiply this by 6 which gives 96 and add it to the other 4*4 = 16 and this also add it to the 4.
To do that question hellion you must first integrate the expression to find the integral of the expression since we are dealing in finding distance then subsitute t (4) into the expression.
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11 comments:
Before we get into the questions, anybody know how to integrate a polynomial???? I tink we have to add 1 to the power and divide by the power. for example
find the integral of 2x^5.
If we add 1 to the power we wud get 6, and divide by the power which is 6, so we wud get 2/6x^6 + c. ENT????
yes coke baby your ethod is correct. Now Solve:
The velocity of a car is given by the equation v = 6t^2 + 4t + t.
Find the distance the car travels after a time, t= 4.
well for this question:
i think that yuh hav to integrate to find the area under the curve after the time = 4.
therefore: the integral answer wud be:
2t^3+ 2t^2 +(t^2/2) + c
and plug in t=4
= 128 + 32 + 8 = 168units
A particle moves along the x-axis with acceleration a = 2t - 3 m/s^2. At time t = 0, it is at the origin and moving with a speed of 4 m/s in the positive direction. Find formulas for its velocity v and position s, and determine where it changes direction and where it is moving to the left.
first integration is the total opposite of defrentiation, like (+) and (-). Hence when u defrentiate y=4x^2 and u get
:dy/dx = 8x,
and now if you integrate;dy/dx = 8x
you will get :y=4x^2.
MISS!! with respect ot your question, we have to keep in mind
v = ds/dt and a = dv/dt.
Therefore dv/dt = 2t - 3 m/s^2 so to get v we must integrate dv/dt.
we wud get v = t^2 - 3t + c. EVERYBODY KNOW HOW TO INTEGRATE RIGHT..SHOCKS I HAVE TO GO AND MEET A GIRL NOW. IL GET BACK TO U ALL LATER... i
hey hellion i think i wud give your quest. a try but correct me if i am wrong
well first of all i think you will have to substitute where t is put for and solve.
so it will be
6t^2 + 4t + t. where t=4
6(4)^2 + 4(4)+ 4
remeber your BODMAS rule befor you go any further
solve brackets first
so it will be 4*4whichwas really the 4 squared and this equals 16 then multiply this by 6 which gives 96 and add it to the other 4*4 = 16 and this also add it to the 4.
96+16+4=116 m
I think thats the distance he will cover
You will now end up with
64 +
intergration is just simply the opposte of differentiation
instead of multiplying by the power we divide
instead of decreasing the power by one we increase it by one
e.g
8x^3 + 23x + x^2 dx
8x^4/4 + 23x^2/2 + x^3/3
Sumbody!! tell me sumting!!! Find the area under the curve
y = x^3 that is between the lines x = 1 and x = 2.
HELLO!!!! ANYONE DERE?????
To do that question hellion you must first integrate the expression to find the integral of the expression since we are dealing in finding distance then subsitute t (4) into the expression.
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