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Q: Integral of 3 raised to x?

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integral (a^x) dx = (a^x) / ln(a)

This integral cannot be performed analytically. Ony when the integral is taken from 0 to infinity can it be computed by squaring the integral and applying a change of variable (switching to polar coordinates). if desired I could show how to do this.

better place to ask would be yahoo answers

The general form is (2/3)*x^(3/2) + C

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

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integral (a^x) dx = (a^x) / ln(a)

(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

maths signs

A primitive to e^(x^(1/3)) is (e^(x^(1/3)))*(6-6x^(1/3)+3x^(2/3))

(e^x)^8 can be written as e^(8*x), so the integral of e^(8*x) = (e^(8*x))/8 or e8x/ 8, then of course you have to add a constant, C.

I will assume that this is sopposed to be integrated with respect to x. To make this problem easier, imagine that the integrand is x raised to the negative 3. The integral is 1/(-2x-2) plus some constant c.

x^3 + C

The integral of -x2 is -1/3 x3 .

Using information from the Wolframalpha site. It seems that this integral can't be expressed as a finite amount of standard functions; you can go to the Wolfram Alpha site, and type "integral x^x", to get a series expansion if you are interested.

The integral would be 10e(1/10)x+c

You add one to the power, and then divide by the power that it has so you would have: Integral of x = (x^2)/2 Integral of x^2 = (x^3)/3 Etc.

This integral cannot be performed analytically. Ony when the integral is taken from 0 to infinity can it be computed by squaring the integral and applying a change of variable (switching to polar coordinates). if desired I could show how to do this.

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