0
It depends whether you mean the indefinite integral (also known as the antiderivative), or the definite integral. In initial calculus courses, you usually start with the indefinite integral.In any case, there is no quick way to explain this; several chapters of calculus books are dedicated to learning several different methods to solve integrals, and those methods don't work in all cases.
In general, you need to go through a calculus course, or book, and learn those methods.
There are many techniques, integration by parts being one of the ones most commonly used. To get a good answer you would need to ask about how to integrate a specific function.
Add your answer:
Earn +20 pts
How do you solve the limits of a definite integral?
By using the fundamental theorem of Calculus. i.e. The integral of f(x) = F(x), your limits are [a,b]. Solve: F(b) - F(a). The FTC, second part, says that if f is a continuous real valued function of [a,b] then the integral from a to b of f(x)= F(b) - F(a) where F is any antiderivative of f, that is, a function such that F'(x) = f(x). Example: Evaluate the integral form -2 to 3 of x^2. The integral form -2 to 3 of x^2 = F(-2) - F(3) = -2^3/3 - 3^3/3 = -8/3 - 27/3 = -35/3
How do you evaluate definite integrals?
In order to evaluate a definite integral first find the indefinite integral. Then subtract the integral evaluated at the bottom number (usually the left endpoint) from the integral evaluated at the top number (usually the right endpoint). For example, if I wanted the integral of x from 1 to 2 (written with 1 on the bottom and 2 on the top) I would first evaluate the integral: the integral of x is (x^2)/2 Then I would subtract the integral evaluated at 1 from the integral evaluated at 2: (2^2)/2-(1^2)/2 = 2-1/2 =3/2.
Integral formula of a raised to x?
integral (a^x) dx = (a^x) / ln(a)
Integral of 1 divided by sinx cosx?
Integral of [1/(sin x cos x) dx] (substitute sin2 x + cos2 x for 1)= Integral of [(sin2 x + cos2 x)/(sin x cos x) dx]= Integral of [sin2 x/(sin x cos x) dx] + Integral of [cos2 x/(sin x cos x) dx]= Integral of (sin x/cos x dx) + Integral of (cos x/sin x dx)= Integral of tan x dx + Integral of cot x dx= ln |sec x| + ln |sin x| + C
Integral of e to the power of x?
The antiderivative, or indefinite integral, of ex, is ex + C.
How do you solve 5.4e0.06t using the fundamental theorem of calculus?
We need more information. Is there a limit or integral? The theorem states that the deivitive of an integral of a function is the function
How I solve the limit of the definite integral?
http://www.math.ucdavis.edu/~kouba/CalcTwoDIRECTORY/defintdirectory/DefInt.html
What is the integral of 0?
The integral of 0 is some constant C. You can solve for this constant by using boundry conditions if there are any given; otherwise, just put C.
How do you solve plane area of integral calculus?
find the area of bounded by the two curves. y=9-x
How do you solve integral problems?
That depends on the specific problem. There is no simple procedure that will let you systematically solve any integral. There are even integrals that look fairly simple (such as the integral of (sin x) / x, if I remember correctly), which can't be solved as a finite number of elementary functions. You need to learn different tricks and techniques (taught in calculus books), know when you can apply each one, and quite often, experiment.
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 are Line integral Surface integral and Volume integral in simple words?
A line integral is a simple integral. they look like: integral x=a to b of (f(x)). A surface integral is an integral of two variables. they look like: integral x=a to b, y=c to d of (f(x,y)). or integral x=a to b of (integral y=c to d of (f(x,y))). The second form is the nested form. A pair of line integrals, one inside the other. This is the easiest way to understand surface integrals, and, normally, solve surface integrals. A volume integral is an integral of three variables. they look like: integral x=a to b, y=c to d, z=e to f of (f(x,y,z)). or integral x=a to b of (integral y=c to d of (integral z=e to f of f(x,y,z))). the above statement is wrong, the person who wrote this stated the first 2 types of integrals as regular, simple, scalar integrals, when line and surface integrals are actually a form of vector calculus. in the previous answer, it is stated that the integrand is just some funtion of x when it is actually usually a vector field and instead of evaluating the integral from some x a to b, you will actually be evaluating the integral along a curve that you will parametrize to get the upper and lower bounds of the integral. as you can see, these are a lot more complicated. looking at your question tho, i dont think you want the whole expanation on how to solve these problems, but more so what they are and what they are used for, because these can be a pain to solve and there are also several ways to solve them indirectly. line integrals have an important part in physics because they alow us to calculate things such as work that have vector values rather than just scalar values as you can use these integrals to describe a particles path along a curve in a force field. surface integrals help us calculate things like flux, or how fluid flows over a surface. if you want to learn more, look into things like greens theorem, or the divergence theorem. p.s. his definition of a surface integral is acutally how you find the volume of a region
What is the integral of e to the power of tan inverse dx?
This integral is too complex for me to put it here (Suffice it to say it involves the Hypergeometric function and imaginary numbers). Go to http://integrals.wolfram.com/index.jsp?expr=e^(arctan(x))&random=false To solve this and other integrals. It can solve nearly every one that can be solved. Hope this helps!
How do you solve integral with equation having x w as a variables but due to one variable of them like w?
By making w in terms of x
How would you solve the integral of 1 plus tan2x plus tan squared 2x?
Integral of 1 is x Integral of tan(2x) = Integral of [sin(2x)/cos(2x)] =-ln (cos(2x)) /2 Integral of tan^2 (2x) = Integral of sec^2(2x)-1 = tan(2x)/2 - x Combining all, Integral of 1 plus tan(2x) plus tan squared 2x is x-ln(cos(2x))/2 +tan(2x)/2 - x + C = -ln (cos(2x))/2 + tan(2x)/2 + C
Integral lnx2 1?
If you mean integral ln(x2+1)dx, it is 2tan-1(x)+x[ln(x2+1)-2]. To find this and other integrals, go to http://integrals.wolfram.com/index.jsp?expr=ln(x^2%2B1)&random=false It can solve nearly every integral I've encountered.
Integral of sec2x-cosx plus x2dx?
I wasn't entirely sure what you meant, but if the problem was to find the integral of [sec(2x)-cos(x)+x^2]dx, then in order to get the answer you must follow a couple of steps:First you should separate the problem into three parts as you are allowed to with integration. So it becomes the integral of sec(2x) - the integral of cos(x) + the integral of x^2Then solve each part separatelyThe integral of sec(2x) is -(cos(2x)/2)The integral of cos(x) is sin(x)The integral of x^2 isLastly you must combine them together:-(cos(2x)/2) - sin(x) + (x^3)/3
