Math and Arithmetic
Algebra
Calculus

# What is the integral of -e to the -x?

###### Answered 2012-05-18 23:30:32

Know that ∫eu du = eu du/dx + c

∫-e-x dx = e-x + c

But ∫eu du = eu + c

Perhaps we are integrating -(e-x ) though the question might be (-e)-x

Question is not clear.

๐
0
๐คจ
0
๐ฎ
0
๐
0

## Related Questions

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

(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.

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!

The antiderivative, or indefinite integral, of ex, is ex + C.

if you mean e to the x power times log of x, it is e to the x divided by x

Assume the expression is: &int; sin(x)x&sup2;e^x dx Then: Take the integral: integral e^x x^2 sin(x) dx For the integrand e^x x^2 sin(x), integrate by parts, integral f dg = f g- integral g df, where f = x^2, dg = e^x sin(x) dx, df = 2 x dx, g = 1/2 e^x (sin(x)-cos(x)): = 1/2 e^x x^2 sin(x)-1/2 (e^x x^2 cos(x))- integral e^x x (sin(x)-cos(x)) dx Expanding the integrand e^x x (sin(x)-cos(x)) gives e^x x sin(x)-e^x x cos(x): = 1/2 e^x x^2 sin(x)-1/2 (e^x x^2 cos(x))- integral (e^x x sin(x)-e^x x cos(x)) dx Integrate the sum term by term and factor out constants: = 1/2 e^x x^2 sin(x)-1/2 (e^x x^2 cos(x))- integral e^x x sin(x) dx+ integral e^x x cos(x) dx For the integrand e^x x sin(x), integrate by parts, integral f dg = f g- integral g df, where f = x, dg = e^x sin(x) dx, df = dx, g = 1/2 e^x (sin(x)-cos(x)): = 1/2 e^x x^2 sin(x)-1/2 e^x x^2 cos(x)-1/2 e^x x sin(x)+1/2 e^x x cos(x)+ integral e^x x cos(x) dx+1/2 integral e^x (sin(x)-cos(x)) dx Expanding the integrand e^x (sin(x)-cos(x)) gives e^x sin(x)-e^x cos(x): = 1/2 e^x x^2 sin(x)-1/2 e^x x^2 cos(x)-1/2 e^x x sin(x)+1/2 e^x x cos(x)+ integral e^x x cos(x) dx+1/2 integral (e^x sin(x)-e^x cos(x)) dx Integrate the sum term by term and factor out constants: = 1/2 e^x x^2 sin(x)-1/2 e^x x^2 cos(x)-1/2 e^x x sin(x)+1/2 e^x x cos(x)+1/2 integral e^x sin(x) dx-1/2 integral e^x cos(x) dx+ integral e^x x cos(x) dx For the integrand e^x cos(x), use the formula integral exp(alpha x) cos(beta x) dx = (exp(alpha x) (alpha cos(beta x)+beta sin(beta x)))/(alpha^2+beta^2): = 1/2 e^x x^2 sin(x)-1/2 e^x x^2 cos(x)-1/4 e^x sin(x)-1/2 e^x x sin(x)-1/4 (e^x cos(x))+1/2 e^x x cos(x)+1/2 integral e^x sin(x) dx+ integral e^x x cos(x) dx For the integrand e^x sin(x), use the formula integral exp(alpha x) sin(beta x) dx = (exp(alpha x) (alpha sin(beta x)-beta cos(beta x)))/(alpha^2+beta^2): = 1/2 e^x x^2 sin(x)-1/2 e^x x^2 cos(x)-1/2 e^x x sin(x)-1/2 (e^x cos(x))+1/2 e^x x cos(x)+ integral e^x x cos(x) dx For the integrand e^x x cos(x), integrate by parts, integral f dg = f g- integral g df, where f = x, dg = e^x cos(x) dx, df = dx, g = 1/2 e^x (sin(x)+cos(x)): = 1/2 e^x x^2 sin(x)-1/2 e^x x^2 cos(x)+e^x x cos(x)-1/2 e^x cos(x)-1/2 integral e^x (sin(x)+cos(x)) dx Expanding the integrand e^x (sin(x)+cos(x)) gives e^x sin(x)+e^x cos(x): = 1/2 e^x x^2 sin(x)-1/2 e^x x^2 cos(x)+e^x x cos(x)-1/2 e^x cos(x)-1/2 integral (e^x sin(x)+e^x cos(x)) dx Integrate the sum term by term: = 1/2 e^x x^2 sin(x)-1/2 e^x x^2 cos(x)+e^x x cos(x)-1/2 e^x cos(x)-1/2 integral e^x sin(x) dx-1/2 integral e^x cos(x) dx For the integrand e^x cos(x), use the formula integral exp(alpha x) cos(beta x) dx = (exp(alpha x) (alpha cos(beta x)+beta sin(beta x)))/(alpha^2+beta^2): = 1/2 e^x x^2 sin(x)-1/2 e^x x^2 cos(x)-1/4 e^x sin(x)+e^x x cos(x)+-3/4 e^x cos(x)-1/2 integral e^x sin(x) dx For the integrand e^x sin(x), use the formula integral exp(alpha x) sin(beta x) dx = (exp(alpha x) (alpha sin(beta x)-beta cos(beta x)))/(alpha^2+beta^2): = 1/2 e^x x^2 sin(x)-1/2 e^x x^2 cos(x)-1/2 e^x sin(x)+e^x x cos(x)-1/2 e^x cos(x)+constant Which is equal to: Answer: | | = 1/2 e^x ((x^2-1) sin(x)-(x-1)^2 cos(x))+constant

Take the integral: integral e^x x^2 sin(x) dx For the integrand e^x x^2 sin(x), integrate by parts, integral f dg = f g- integral g df, where f = x^2, dg = e^x sin(x) dx, df = 2 x dx, g = 1/2 e^x (sin(x)-cos(x)): = 1/2 e^x x^2 sin(x)-1/2 (e^x x^2 cos(x))- integral e^x x (sin(x)-cos(x)) dx Expanding the integrand e^x x (sin(x)-cos(x)) gives e^x x sin(x)-e^x x cos(x): = 1/2 e^x x^2 sin(x)-1/2 (e^x x^2 cos(x))- integral (e^x x sin(x)-e^x x cos(x)) dx Integrate the sum term by term and factor out constants: = 1/2 e^x x^2 sin(x)-1/2 (e^x x^2 cos(x))- integral e^x x sin(x) dx+ integral e^x x cos(x) dx For the integrand e^x x sin(x), integrate by parts, integral f dg = f g- integral g df, where f = x, dg = e^x sin(x) dx, df = dx, g = 1/2 e^x (sin(x)-cos(x)): = 1/2 e^x x^2 sin(x)-1/2 e^x x^2 cos(x)-1/2 e^x x sin(x)+1/2 e^x x cos(x)+ integral e^x x cos(x) dx+1/2 integral e^x (sin(x)-cos(x)) dx Expanding the integrand e^x (sin(x)-cos(x)) gives e^x sin(x)-e^x cos(x): = 1/2 e^x x^2 sin(x)-1/2 e^x x^2 cos(x)-1/2 e^x x sin(x)+1/2 e^x x cos(x)+ integral e^x x cos(x) dx+1/2 integral (e^x sin(x)-e^x cos(x)) dx Integrate the sum term by term and factor out constants: = 1/2 e^x x^2 sin(x)-1/2 e^x x^2 cos(x)-1/2 e^x x sin(x)+1/2 e^x x cos(x)+1/2 integral e^x sin(x) dx-1/2 integral e^x cos(x) dx+ integral e^x x cos(x) dx For the integrand e^x cos(x), use the formula integral exp(alpha x) cos(beta x) dx = (exp(alpha x) (alpha cos(beta x)+beta sin(beta x)))/(alpha^2+beta^2): = 1/2 e^x x^2 sin(x)-1/2 e^x x^2 cos(x)-1/4 e^x sin(x)-1/2 e^x x sin(x)-1/4 (e^x cos(x))+1/2 e^x x cos(x)+1/2 integral e^x sin(x) dx+ integral e^x x cos(x) dx For the integrand e^x sin(x), use the formula integral exp(alpha x) sin(beta x) dx = (exp(alpha x) (alpha sin(beta x)-beta cos(beta x)))/(alpha^2+beta^2): = 1/2 e^x x^2 sin(x)-1/2 e^x x^2 cos(x)-1/2 e^x x sin(x)-1/2 (e^x cos(x))+1/2 e^x x cos(x)+ integral e^x x cos(x) dx For the integrand e^x x cos(x), integrate by parts, integral f dg = f g- integral g df, where f = x, dg = e^x cos(x) dx, df = dx, g = 1/2 e^x (sin(x)+cos(x)): = 1/2 e^x x^2 sin(x)-1/2 e^x x^2 cos(x)+e^x x cos(x)-1/2 e^x cos(x)-1/2 integral e^x (sin(x)+cos(x)) dx Expanding the integrand e^x (sin(x)+cos(x)) gives e^x sin(x)+e^x cos(x): = 1/2 e^x x^2 sin(x)-1/2 e^x x^2 cos(x)+e^x x cos(x)-1/2 e^x cos(x)-1/2 integral (e^x sin(x)+e^x cos(x)) dx Integrate the sum term by term: = 1/2 e^x x^2 sin(x)-1/2 e^x x^2 cos(x)+e^x x cos(x)-1/2 e^x cos(x)-1/2 integral e^x sin(x) dx-1/2 integral e^x cos(x) dx For the integrand e^x cos(x), use the formula integral exp(alpha x) cos(beta x) dx = (exp(alpha x) (alpha cos(beta x)+beta sin(beta x)))/(alpha^2+beta^2): = 1/2 e^x x^2 sin(x)-1/2 e^x x^2 cos(x)-1/4 e^x sin(x)+e^x x cos(x)+-3/4 e^x cos(x)-1/2 integral e^x sin(x) dx For the integrand e^x sin(x), use the formula integral exp(alpha x) sin(beta x) dx = (exp(alpha x) (alpha sin(beta x)-beta cos(beta x)))/(alpha^2+beta^2): = 1/2 e^x x^2 sin(x)-1/2 e^x x^2 cos(x)-1/2 e^x sin(x)+e^x x cos(x)-1/2 e^x cos(x)+constant Which is equal to: Answer: | | = 1/2 e^x ((x^2-1) sin(x)-(x-1)^2 cos(x))+constant

(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)&amp;random=false

let u = x du=dx let dv= e^x v=e^x &int; xe^(x)dx = xe^x - &int; e^(x)dx = xe^x - e^x = e^x ( x-1 ) + c

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.

-e^(-x) or negative e to the negative x this is because you multiply the function (e) by: 1 / (the derivative of the power ... in this case: -1) e^(-x) * (1/-1) = -e^(-x) Don't forget to add your constant!

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

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

If the probability distribution function for the random variable X is f(x), then first calculate E(X) = integral of x*f(x)dx over the whole real line. Noxt calculate E(X2) = integral of x2*f(x)dx over the whole real line. Then Variance(X) = E(X2) - [E(X)]2 and finally, SD(X) = sqrt[Variance(X)].

dy/dx = 3^x * ln(3)integral = (3^x) / ln(3)To obtain the above integral...Let y = 3^xln y = x ln 3y = e^(x ln 3)(i.e. 3^x is the same as e^(x ln 3) ).The integral will then be 3^x / ln 3 (from linear composite rule and substitution after integration).

Evaluate the integral? Use integration by parts. uv - int v du u = e^x du = e^x dv = sinx v = -cosx int e^x sinx dx -e^x cosX - int -cosx e^x -e^x cosx + sinx e^x + C ----------------------------------

&acirc;&circ;&laquo; ex dx = ex + CC is the constant of integration.

The integral of x cos(x) dx is cos(x) + x sin(x) + C

If a random variable X has mean value m, and standard deviation s, then Skewness = E{[(x - m)/s]3} which can be simplified to skewness = [E(X3) - 3ms2 - m3]/s3 and for discrete X, E(X3) = sum of x3*Prob(X = x) where the summation is over all possible values of x. While for continuous X, E(X3) = integral of x3*f(x) where the integral is over the domain of X.

In reimann stieltjes integral if we assume a(x) = x then it becomes reimann integral so we can say R-S integral is generalized form of reimann integral.

The integral of 2-x = 2x - (1/2)x2 + C.

###### CalculusMath and ArithmeticSchool SubjectsProbabilityScience

Copyright ยฉ 2021 Multiply Media, LLC. All Rights Reserved. The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply.