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What else can I help you with?
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!!!!
Integrate square root of 4 xdivide by x?
∫[√(4x) / x] dx = ∫(2 / √x)dx = 2∫(x-1/2) dx = 2(2x1/2 + C) = 4√x + C
Integrate x 5x dx?
integrate(x5x dx) simplifies to integrate(5x^2 dx), and using the power rule of integration, add one to the power of x and divide the term by that number. Thus, x5x dx integrated is (5/3)x^3
How do you integrate cot3x dx?
1/3ln(sin3x) + C
How do you integrate 1 over x-1?
2
How do you calculate average output voltage for a square wave?
"Average" can mean many things. If what you're meaning is the root mean square, you can calculate the same as for a sine wave: integrate the square of the waveshape from 0 to the end of the first period, divide by the length of the period, and take the square root of this value. For a 50% square wave (it's at amplitude a for 50% of the cycle, and at 0 for the other 50%, so the integral will be over only 1/2 the period): sqrt[1/P * integral (a^2) dx, from 0 to 1/2*P] = sqrt[a^2/P * (x from 0 to 1/2*P)] sqrt[a^2 / P * (P/2) = a/sqrt(2) For 33.3%, integrate over 1/3 of the period, so RMS = a / sqrt(3).
How you integrate xxxx 1 dx?
.2x^5+x+C
What is the derivative function for square root 1-2x?
3
How do you Integrate the following S4x3-2x2 x-1dx in Calculus?
∫(4x3 - 2x2 + x - 1) dx You can integrate this by taking the antiderivative of each term. Each of these terms is in the format axn, the antiderivative of which is axn-1/n: = ∫(4x3)dx - ∫(2x2)dx + ∫(x)dx - ∫(1)dx = x4 - 2x3/3 + x2/2 - x + C
What is the integration of cotx?
The integral of cot (x) dx is ln (absolute value (sin (x))) + C. Without using the absolute value, you can use the square root of the square, i.e. ln (square root (sin2x)) + C
How do you integrate f dx?
f dx ?? do you mean f df ? int(f df) (1/2)f2 + C --------------
How do you integrate cotxdx?
The integral of cot(x)dx is ln|sin(x)| + C
How do you integrate piecewise continuous functions?
Simply integrate all the pieces apart, en add them up. This is allowed, because int_a^c f(x)dx = int_a^b f(x)dx + int_b^c f(x)dx for all a,b,c in dom(f).
