F = t ⇨ df = dt
dg = sec² 2t dt ⇨ g = (1/2) tan 2t
⇔
integral of t sec² 2t dt = (1/2) t tan 2t - (1/2) integral of tan 2t dt
u = 2t ⇨ du = 2 dt
As integral of tan u = - ln (cos (u)), you get :
integral of t sec² 2t dt = (1/4) ln (cos (u)) + (1/2) t tan 2t + constant
integral of t sec² 2t dt = (1/2) t tan 2t + (1/4) ln (cos (2t)) + constant
integral of t sec² 2t dt = (1/4) (2t tan 2t + ln (cos (2t))) + constant ⇦ answer
Answer:
The first one
Step-by-step explanation:
304,913
Answer:
I dunno what you wrote at the top, but I solved the picture
Step-by-step explanation:
Answer:
Step-by-step explanation:
b it is b just do what u think is right because i think u can do it no matter what
There is no hard and fast rule to select the class width. It largely depends on our application.However, one thing that should be kept in mind is that the number of classes should neither to be too low nor too high. So keeping this thing in mind, the class width is select.
The range of the data is = Maximum- Minimum = 96 - 11 = 85
10 classes will be most suited for this data.
The class width for each data can be calculated as:
Class Width = Range / Number of Classes = 85/10 = 8.5
Class width is always rounded to nearest next integer. So the class width will be 9 in this case.
So, the best value of class width or interval width for the given data will be 9.
