I think you meant the integral of 21-x dx, or as I write it ,/`21-x dx.
Use u-substitution where u = 1 - x. Note that the derivative of this equation is du = -1 dx, so we know dx = - du.
Rewriting the original integral in terms of u, integrating, then putting it back in terms of x gives us:
,/` 21-x dx
,/` 2u (-du)
- ,/` 2u du
- 2u / ln(2)
- 21-x / ln(2)
Note that the second to last line can be found in any integral table, and ln denotes natural log.
Now the bounds go from x = 0 to x = t, so:
[- 21-t / ln(2)] - [- 21-0 / ln(2)]
- 21-t / ln(2) + 2 / ln(2)
(-21-t + 2)/ln(2)
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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).
Let y = sin 2x Then dy/dx = 2*cos 2x Also, when x = 0, y = 0 and when x = pi/6, y = sin(2*pi/6) = sin(pi/3) = sqrt(3)/2 Therefore, the integral becomes definite integral from 0 to √3/2 of 1/2*y3dy = difference between 1/2*(y4)/4 = (y4)/8 evaluated at √3/2 and 0. = (√3/2)4 /8 = 9/128 = 0.0703 approx.
0.
e3.14?-The derivative of eu is:d/dx(eu)=eu*[d/dx(u)]d/dx(e3.14)=e3.14*[d/dx(3.14)]d/dx(e3.14)=e3.14*(0) ;The derivative of a constant is 0.d/dx(e3.14)=0
y = x2 - 5xThe integral of [ y dx ] = x3/3 - 5x2/2 + CAt x=0, the integral = 0At x=5, the integral = 125/3 - 125/2 = -20.8333At x=6, the integral = 72 - 90 = -18At x = -2, the integral = -8/3 - 10 = -12.6666a). from 0 to 5, the area is -20.8333 (20.8333 below the x-axis)b). from 0 to 6, the area is -18.c). from -2 to +5, the area is (-20.8333 + 12.6666) = +9.8333