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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).
The integral for csc(u)dx is -ln|csc(u) + cot(u)| + C.
This can easily be integrated by substitution. Integrate[(6x3+8)418x2 dx] let u=(6x3+8) du/dx= 18x2 du=18x2 dx Substituting u in gives... Integrate [u4 du] = (1/5)(u5) + C Substituting x back in gives.. (1/5)(6x3+8)5 + C
Assuming you are integrating with respect to one of the three variables, you integrate normally. For example: ∫(x+y+z)dx = ∫ x dx + ∫ y dx + ∫ z dx (Integral of the sum is the sum of the integrals) = x^2/x + yx + zx + C Or a harder one: ∫ (sin^2(y)+sqrt(z))/x dx = (sin^2(y) + sqrt(z))*∫ 1/x dx (Factor out constants) = ln(x)*(sin^2(y) + sqrt(z)) tl;dr: just do it normally with normal integration rules
∫[√(4x) / x] dx = ∫(2 / √x)dx = 2∫(x-1/2) dx = 2(2x1/2 + C) = 4√x + C
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
1/3ln(sin3x) + C
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"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).
.2x^5+x+C
∫(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
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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
f dx ?? do you mean f df ? int(f df) (1/2)f2 + C --------------
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).
The integral of cot(x)dx is ln|sin(x)| + C