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What is the relation between definite integrals and areas?
Consider the integral of sin x over the interval from 0 to 2pi. In this interval the value of sin x rises from 0 to 1 then falls through 0 to -1 and then rises again to 0. In other words the part of the sin x function between 0 and pi is 'above' the axis and the part between pi and 2pi is 'below' the axis. The value of this integral is zero because although the areas enclosed by the parts of the function between 0 and pi and pi and 2pi are the same the integral of the latter part is negative. The point I am trying to make is that a definite integral gives the area between a function and the horizontal axis but areas below the axis are negative. The integral of sin x over the interval from 0 to pi is 2. The integral of six x over the interval from pi to 2pi is -2.
How do you use an arc length integral to show the length of the circle of radius r?
The integral from 0 to 2 pi of your constant value r will equal the circumference. This will be equal to 2*pi*r. This can be derived because of the following: Arc length = integral from a to b of sqrt(r^2-(dr/dtheta)^2) dtheta. By substituting the equation r = a constant c, dr/dtheta will equal 0, a will equal 0, and b will equal 2pi (the radians in a circle). By substitution, this becomes the integral from 0 to 2 pi of sqrt(c^2 + 0)dtheta, which leads us back to the original formula.
What is the integral of mod x function?
The integral of the absolute value function, |x|, is given by the piecewise function ∫|x|dx = (x^2)/2 + C for x ≥ 0 and ∫|x|dx = (-x^2)/2 + C for x < 0, where C is the constant of integration. This is because the absolute value function changes its behavior at x = 0, resulting in two different expressions for the integral depending on the sign of x.
What is the indefinite integral of 3sinx-5cosx?
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What is the convulusion integral?
Do you mean the Convolution Integral?
