0
2
Still curious? Ask our experts.
Chat with our AI personalities
Coach
Ezra
Vivi
Add your answer:
Earn +20 pts
How do you integrate 1 over x-1?
2
What is the integral of 2-2x with limits 0 to t?
Integrate(0->t) (2-2x) dx is the integral correct (Integrate(0->t) (2-2x) dy would be different, you must state integrate with respect to what otherwise it can be anything) So integrals preserves sum and product with constants. i.e. Integrate (2-2x) dx = Integrate 2 dx - 2integrate x dx = 2x - x^2 + C By Fundamental Theorem of Calculus, take any anti-derivative, say C = 0 would be fine, and Integral(0->t)(2-2x) dx = (2x-x^2)|(0->t) = (2t-t^2) - 0 = 2t-t^2 It is a special case of the Second Fundamental Theorem of Calculus -- integral(0->t) f(x) dx is an anti-derivative of f(x).
What is the derivative of y equals square root of x when x is positive?
y = x^(1/2) (NB power of '1/2' mean the 'square root'. Hence dy/dx = (1/2)x^(-1/2) or dy/dx = 1/ [2x^(1/2)]
How do you integrate the cscnx?
The integral for csc(u)dx is -ln|csc(u) + cot(u)| + C.
What is the derivative of square root of x-5?
Use Cahin Rule dy/dx = dy/du X du/dx Hence y = ( x-5)^(1/2) Let x - 5 = u Hence dy/du = (1/2)u^(-1/2) du/dx = 1 Combining dy/dx = (1/2)(x-5)^(-1/2)) X 1 dy/dx = 1/ [2(x-5)^(1/2)] Done!!!!
