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.
The integral would be 10e(1/10)x+c
(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
integral of e to the power -x is -e to the power -x
C = k*a*d*e^3/sqrt(m) where k is a constant.
replace square root o x with t.
better place to ask would be yahoo answers
maths signs
(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.
2 is the cube root of eight
2 = Cube Root of Eight
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.
45
Cube root is the same as to the power of a third; when multiplying/dividing powers of a number add/subtract the powers; when a power is to another power, multiply the powers; as it is all e to some power: e³/(e²)^(1/3) × e^13 = e³/e^(2/3) × e^13 = e^(3 - 2/3 + 13) = e^(15 1/3) = e^(46/3) Which can also be expressed as "the cube root of (e to the power 46)" or "(the cube root of e) to the power 46".
The integral would be 10e(1/10)x+c
The square root of 16 is 4 and 4 cubed is 16x4=64
The integral you are referring to is the normalization of a Gaussian function. Specifically, the integral of ( e^{-5x^2} ) from negative infinity to positive infinity equals ( \sqrt{\frac{\pi}{5}} ). Therefore, multiplying that result by ( 7 ) gives ( 7 \sqrt{\frac{\pi}{5}} ). The final expression is ( 7 \sqrt{\frac{\pi}{5}} ).