how do you integrate (x^3)(e^-3x)dx
d/dx(x^3e^-3x) = (x^3)(-3e^-3x) + 3x^2(e^-3x)
Is this correct and how do you continue from here?
You do it by parts, several times. https://www.integral-calculator.com/ can work it into steps to show you how....it is a busy tablet of work.
let
u = x^n
du = nx^(n-1) dx
dv = e^(ax)
v = 1/a e^(ax)
so,
∫x^n e^(ax) dx = 1/a x^n e^(ax) - n/a ∫x^(n-1) e^(ax) dx
So, to start off,
∫(x^3)(e^-3x)dx = 1/-3 x^3 e^(-3x) - (3/-3)∫x^2 e^(-3x) dx
= -1/3 x^3 e^(-3x) + ∫x^2 e^(-3x) dx
and so on till the x stuff is gone
Yes, your differentiation is correct.
To integrate the function (x^3)(e^-3x) dx, you can use integration by parts. The formula for integration by parts is:
∫ (u dv) = uv - ∫ (v du)
Let's assign u and dv as follows:
u = x^3
dv = e^-3x dx
Taking the derivatives and integrating:
du = 3x^2 dx
v = ∫ e^-3x dx = (-1/3) e^-3x
Now we can substitute these values into the formula:
∫ (x^3)(e^-3x) dx = uv - ∫ (v du)
= (x^3)(-1/3)(e^-3x) - ∫ ((-1/3)e^-3x)(3x^2) dx
= (-1/3)(x^3)(e^-3x) + (1/3)∫ (3x^2)(e^-3x) dx
Now we have (∫ (3x^2)(e^-3x) dx) left to solve. We can use integration by parts again.
Let's assign u and dv as follows:
u = 3x^2
dv = e^-3x dx
Taking the derivatives and integrating:
du = 6x dx
v = ∫ e^-3x dx = (-1/3) e^-3x
Now we substitute these values into the formula for the second integration by parts:
∫ (3x^2)(e^-3x) dx = uv - ∫ (v du)
= (3x^2)(-1/3)(e^-3x) - ∫ ((-1/3)e^-3x)(6x) dx
= (-x^2)(e^-3x) + (2/3)∫ (xe^-3x) dx
Now we have (∫ (xe^-3x) dx) left to solve. We can use integration by parts again.
Let's assign u and dv as follows:
u = x
dv = e^-3x dx
Taking the derivatives and integrating:
du = dx
v = ∫ e^-3x dx = (-1/3) e^-3x
Now we substitute these values into the formula for the third integration by parts:
∫ (xe^-3x) dx = uv - ∫ (v du)
= (x)(-1/3)(e^-3x) - ∫ ((-1/3)e^-3x)(1) dx
= (-x/3)(e^-3x) + (1/3)∫ e^-3x dx
= (-x/3)(e^-3x) + (1/3)(-1/3)(e^-3x)
We can simplify this to:
= (-x/3)(e^-3x) - (1/9)(e^-3x)
So, the final result of the integral ∫ (x^3)(e^-3x) dx is:
(-1/3)(x^3)(e^-3x) + (1/3)(-x/3)(e^-3x) - (1/9)(e^-3x) + C
where C is the constant of integration.
Yes, your first step is correct. You found the derivative of the function (x^3)(e^-3x) correctly, which is (x^3)(-3e^-3x) + 3x^2(e^-3x).
To integrate the function (x^3)(e^-3x), you need to reverse the process of differentiation.
Notice that the expression (-3e^-3x) in the derivative is the derivative of the exponential function e^-3x. This indicates that e^-3x is the function you should integrate next.
Integrating e^-3x is straightforward. The integral of e^-3x with respect to x is (-1/3)e^-3x.
Now, let's integrate the other part of the derivative, which is 3x^2(e^-3x). To integrate this term, we use the power rule of integration.
According to the power rule, the integral of x^n with respect to x, when n is not equal to -1, is (x^(n+1))/(n+1). Applying this rule, the integral of x^2 with respect to x is (x^3)/3.
Multiplying this by e^-3x, we get (x^3)(e^-3x)/3.
Therefore, the integral of (x^3)(e^-3x)dx is the sum of these two integrals:
∫(x^3)(e^-3x)dx = (-1/3)e^-3x + (x^3)(e^-3x)/3 + C,
where C is the constant of integration.