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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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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 definite integral from 0 to pi divided by 6 of sin cubed 2x cos 2x dx?
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.
What is -1x plus x?
0.
What is the derivative of e3.14?
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
What is the area of the region between y equals x2 -5x and the x axis a from 0 to5 b from 0 to 6 cfrom-2 to5?
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
What is the integral of 2 times 1-x dx with limits 0 to t?
,/` 2(1 - x) dx,/` 2 - 2x dx2x - x2 ...evaluated from 0 to t gives us...2t - t2 - [2(0) - (0)2]2t - t2
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 integretion of modxdx?
mod x, or |x| is actually a conjunction of two functions: 1) x = -x, for x < 0 2) x = x, for x >= 0. Whenever you're calculating integral of |x|, you have to consider those two functions, for example: integral of |x| from -5 to 4 by dx is a sum of integrals of -x from -5 to 0 by dx and integral of x from 0 to 4 by dx.
What is the integral of 1 divided by x squared?
The indefinite integral of (1/x^2)*dx is -1/x+C.
What is the integral from 0 to pi over 6 sine 2x dx?
Integral from 0 to pi 6sin2xdx: integral of 6sin2xdx (-3)cos2x+c. (-3)cos(2 x pi) - (-3)cos(2 x 0) -3 - -3 0
What is integral of 2x?
The integral of 2x is x^2+c, where c is a constant. If this is a definite integral, meaning that the limits of integration are known, then c=0. If this is an indefinite integral, meaning the limits of integration are unknown, then c should either be left as is or solved for using an initial condition.
What is the integral of -3x?
In order to compute that integral, we need to use the power rule: ∫ xⁿ dx = xn + 1/(n + 1) + c where n is any constant except 0 and -1. Apply that rule to get: ∫ 3x dx = 3 ∫ x dx [Factor out the constant] = 3 ∫ x1 dx [Make note of the exponent] = 3x1 + 1/(1 + 1) + c = 3x2/2 + c So that is the integral of 3x.
How can you graph x -x on a number line if there are no numbers?
Hi, X - X is 0 because any number minus it's self = 0. So the graph of y = x - x is the same as y = 0 If it had been y = x + x then you would add the x's so it would really be y = 2x. My question as how to find ∫ (xx) dx. If anybody knows how to do this please contact me at RapidPixel@gmail.com the integral of x^x dx. the integral of x to the power of x dx
What is a set of example for practice distinguished from theory application or use as of knowledge or skills?
integral form 0 to infinty (x^2) / (e^x+1) dx = ?
What is the definite integral from 0 to pi divided by 6 of sin cubed 2x cos 2x dx?
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.
Why derivative of mod x does not exist graphically?
Because mod(x) is not "smooth at x = 0.Suppose f(x) = mod(x). Then f'(x), if it existed, would be the limit, as dx tends to 0, of [f(x+dx) - f(x)]/dx= limit, as dx tends to o , of [mod(x+dx) - mod(x)]/dxWhen x = 0, this simplifies to mod(dx)/dxIf dx > 0 then f'(x) = -1andif dx < 0 then f'(x) = +1Consequently f'(0) does not exist and hence the derivative of mod(x) does not exist at x = 0.Graphically, it is because at x = 0 the graph is not smooth but has an angle.Because mod(x) is not "smooth at x = 0.Suppose f(x) = mod(x). Then f'(x), if it existed, would be the limit, as dx tends to 0, of [f(x+dx) - f(x)]/dx= limit, as dx tends to o , of [mod(x+dx) - mod(x)]/dxWhen x = 0, this simplifies to mod(dx)/dxIf dx > 0 then f'(x) = -1andif dx < 0 then f'(x) = +1Consequently f'(0) does not exist and hence the derivative of mod(x) does not exist at x = 0.Graphically, it is because at x = 0 the graph is not smooth but has an angle.Because mod(x) is not "smooth at x = 0.Suppose f(x) = mod(x). Then f'(x), if it existed, would be the limit, as dx tends to 0, of [f(x+dx) - f(x)]/dx= limit, as dx tends to o , of [mod(x+dx) - mod(x)]/dxWhen x = 0, this simplifies to mod(dx)/dxIf dx > 0 then f'(x) = -1andif dx < 0 then f'(x) = +1Consequently f'(0) does not exist and hence the derivative of mod(x) does not exist at x = 0.Graphically, it is because at x = 0 the graph is not smooth but has an angle.Because mod(x) is not "smooth at x = 0.Suppose f(x) = mod(x). Then f'(x), if it existed, would be the limit, as dx tends to 0, of [f(x+dx) - f(x)]/dx= limit, as dx tends to o , of [mod(x+dx) - mod(x)]/dxWhen x = 0, this simplifies to mod(dx)/dxIf dx > 0 then f'(x) = -1andif dx < 0 then f'(x) = +1Consequently f'(0) does not exist and hence the derivative of mod(x) does not exist at x = 0.Graphically, it is because at x = 0 the graph is not smooth but has an angle.
How do you find the value of k when the area enclosed by x-axis and the curve y equals root x and the line x equals k is 18?
The area under a curve (ie between the curve and the x-axis) is found by integrating the curve with respect to x: A = ∫ y dx When limits to the integration are given, then the value of the area is the difference between the value of the integral at the limits. The lower limit is where the curve y = √x meets the x-axis, ie at y = 0 → √x = 0 → x = 0 The upper limit is the line x = k → the area is given by ∫ √x dx between 0 and k To integrate √x write it in power format: √x = x^½, giving: ∫ √x dx = ∫ x^½ dx = ⅔x^(3/2) + c The difference between the limits is the area: (⅔k^(3/2) + c) - (⅔0^(3/2) + c) = 18 → ⅔k^(3/2) = 18 → k^(3/2) = 18 ÷ ⅔ = 27 → k = 27^(2/3) = (27^⅓)² = 9 Thus k = 9.
