x=1
integral of e to the power -x is -e to the power -x
I'm not sure if you mean e^x + 17 or e^(x+17) so we'll do both. First, the integral of e^x + 17 because these terms are being added you can integrate them separately: integral((e^x)dx) + integral(17dx) integral of e^x is just e^x + C Integral of 17 is 17x + C, so we get: e^x + 17x + C Second, the integral of e^(x+17) we know how to integrate the form e^u, so just do a u substitution u=x+17 du=dx so we get integral((e^u)du)=e^u + C resubstitute for u and get e^(x+17) + C
The antiderivative, or indefinite integral, of ex, is ex + C.
(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.
maths signs
Writing equations in questions is problematic - some symbols regularly get eliminated.The integral of e to the power x is: e to the power x + C If your expression contains no variables, for example e times e, or e to the power e, then the entire expression is a constant; in this case, the integral is this constant times 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
Use integration by parts. integral of xe^xdx =xe^x-integral of e^xdx. This is xe^x-e^x +C. Check by differentiating. We get x(e^x)+e^x(1)-e^x, which equals xe^x. That's it!
Michael E Lord has written: 'Validation of an invariant embedding method for Fredholm integral equations' -- subject(s): Invariant imbedding, Numerical solutions, Integral equations
The integral of e-2x is -1/2*e-2x + c but I am not sure what "for x0" in the question means.
if you mean e to the x power times log of x, it is e to the x divided by x
Charles E. Simmonds has written: 'The role of differentials and integral calculus in schools'