It is better to integrate the function from 0 to a and then calculate the limit of the result when a goes to 3 from the left because the function is undefined at 3.
We can find integration of a function that is undefined at some point but have a defined area. So the solution in the video is correct. Also if we have to integrate from a to b and graph of the function is discontinuous at a point between a and b suppose c. Then again we can intrgrate it from a to c and from c to b and then we can add these two values to get the answer.
You did not understand me. I did not say that the answer is incorrect, but rather that it lacks precision. We should have replaced the number 3 with a letter such as a, then calculated the integral. After that, we calculate the limit when a approaches 3. An example of this is when calculating the integral of the function f(x)=ln(x) from 1 to 0. If we take your method, we calculate the integral of the function f, which is the function g(x)=xlnx-x. So how do we calculate the integral when the function g is not defined at 0?@@CalculusBooster
You can answer it by sin inverse and it is more faster than trig substitution
It is better to integrate the function from 0 to a and then calculate the limit of the result when a goes to 3 from the left because the function is undefined at 3.
Αυτά είναι στο πανεπιστήμιο;
When we do not use mathematics accurately, we fall into contradictions.
Here is an example f(x)=ln(x) How do you calculate the integral of this function from 1 to 0?
We can find integration of a function that is undefined at some point but have a defined area. So the solution in the video is correct.
Also if we have to integrate from a to b and graph of the function is discontinuous at a point between a and b suppose c. Then again we can intrgrate it from a to c and from c to b and then we can add these two values to get the answer.
You did not understand me. I did not say that the answer is incorrect, but rather that it lacks precision. We should have replaced the number 3 with a letter such as a, then calculated the integral. After that, we calculate the limit when a approaches 3. An example of this is when calculating the integral of the function f(x)=ln(x) from 1 to 0. If we take your method, we calculate the integral of the function f, which is the function g(x)=xlnx-x. So how do we calculate the integral when the function g is not defined at 0?@@CalculusBooster
E'più semplice l'esercizio trasformando il denominatore nella derivata dell'arcsen .
Solution:
3 3 3
∫dx/√(9-x²) = ∫dx/√[9/9*(9-x²)] = ∫dx/{3*√[1/9*(9-x²)]}
0 0 0
3 3
= 1/3*∫dx/√(1-x²/9) = 1/3*∫dx/√[1-(x/3)²] =
0 0
----------------
Substitution: u = x/3 du = 1/3*dx dx = 3*du
upper limit = 1 lower limit = 0
----------------
1
= ∫du/√(1-u²) =
0
----------------
Substitution: u = sin(t) du = cos(t)*dt
upper limit = arcsin(1) = π/2 lower limit = arcsin(0) = 0
----------------
π/2 π/2 π/2 π/2
= ∫cos(t)*dt/√[1-sin²(t)] = ∫cos(t)*dt/cos(t) = ∫dt = [t] = π/2
0 0 0 0