0
Add your answer:
Earn +20 pts
How do you integrate periodic functions?
Same as any other function - but in the case of a definite integral, you can take advantage of the periodicity. For example, assuming that a certain function has a period of pi, and the value of the definite integral from zero to pi is 2, then the integral from zero to 2 x pi is 4.
What is the integral of e raised to x cubed?
(ex)3=e3x, so int[(ex)3dx]=int[e3xdx]=e3x/3 the integral ex^3 involves a complex function useful only to integrations such as this known as the exponential integral, or En(x). The integral is:-(1/3)x*E2/3(-x3). To solve this integral, and for more information on the exponential integral, go to http://integrals.wolfram.com/index.jsp?expr=e^(x^3)&random=false
What is the integral of the derivative with respect to x of a function of x with respect to x?
∫ d/dx f(x) dx = f(x) + C C is the constant of integration.
What is the relationship of integral and differential calculus?
We say function F is an anti derivative, or indefinite integral of f if F' = f. Also, if f has an anti-derivative and is integrable on interval [a, b], then the definite integral of f from a to b is equal to F(b) - F(a) Thirdly, Let F(x) be the definite integral of integrable function f from a to x for all x in [a, b] of f, then F is an anti-derivative of f on [a,b] The definition of indefinite integral as anti-derivative, and the relation of definite integral with anti-derivative, we can conclude that integration and differentiation can be considered as two opposite operations.
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 explain negative integral?
The definite integral of a function: y = f(x) from x = a to x = b is equal to the area between the function curve and the 'x' axis from x = a to 'x' = b.
Why you use even and odd function in mathematics?
If you know that a function is even (or odd), it may simplify the analysis of the function, for several purposes. One example is the calculation of definite integrals: for an odd function, the integral of a function from (-x) to (x) (note 1) is zero; for an even function, this integral is twice the integral of the function from (0) to (x). Note 1: That is, the area under the function; for negative values, this "area" is taken as negative) is
Importance of anti-differentiation in integral calculus?
If F(x) is a function, and F ‘(x) = f(x), then F(x) is the antiderivative (or indefinite integral) of f(x) It is the cornerstone of integral calculus and is used for areas, volumes, lengths and so much more!
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 answer of 1 divide by x square?
What do you mean? As this is a calculus question, I presume that you are asking for a derivative or integral The derivative of any function of the form ƒ(x) = a * x ^ n is ƒ'(x) = a * n * x ^ (n-1) The integral of any function of the form ∫ a*x ^ n is a / (n+1) * x ^ (n+1) + C Your function that you gave is 1 / x^(2) which is equal to: x^(-2) Thus the derivative is: -2 * x^(-3) And the integral is: -x^(-1) + C
What is the integral of a constant multiplied by a function of x with respect to x?
∫ af(x) dx = a ∫ f(x) dx
What is the derivative with respect to x of the integral of a function of x with respect to x?
d/dx ∫ f(x) dx = f(x)
How do you integrate functions?
To integrate a function you find what the function you have is the derivative of. for example the derivative of x^2 is 2x. so the integral of 2x is x^2.
How do you integrate periodic functions?
Same as any other function - but in the case of a definite integral, you can take advantage of the periodicity. For example, assuming that a certain function has a period of pi, and the value of the definite integral from zero to pi is 2, then the integral from zero to 2 x pi is 4.
How you can defferentiate an integral?
If the upper limit is a function of x and the lower limit is a constant, you can differentiate an integral using the Fudamental Theorem of Calculus. For example you can integrate Integral of [1,x^2] sin(t) dt as: sin(x^2) d/dx (x^2) = sin(x^2) (2x) = 2x sin(x^2) The lower limit of integration is 1 ( a constant). The upper limit of integration is a function of x, here x^2. The function being integrated is sin(t)
How can I generate a declining function with constraints on the x and y intercepts so that the integral of the curve is constant?
The integral of a given function between given integration limits will always be a constant. The integral of a given function between variable limits - for example, from 0 to x - can only be a constant if the function is equal to zero everywhere.
What is the second derivative of a function's indefinite integral?
well, the second derivative is the derivative of the first derivative. so, the 2nd derivative of a function's indefinite integral is the derivative of the derivative of the function's indefinite integral. the derivative of a function's indefinite integral is the function, so the 2nd derivative of a function's indefinite integral is the derivative of the function.
