Find the indefinite integral.
∫ x^2(1-x)^6 dx
I am so lost on this. I let u=1-x so du=-dx.
After that, I'm not sure what to do.
go to
http://www.wolframalpha.com/input/?i=integrate+x%2F%281-x%29
type in:
integrate x^2(1-x)^6
It is a serious mess !!!
The only thing I know to do is multiply (1-x)^6 out (use binary coefficients)
then you have a whole string of simple ones
which is why the answer at Wolfram starts with someting times x^9
(1-x)^6
= 1 - 6x + 15x^2 - 20x^3 etc
multiply by x^2
so
x^2 - 6x^3 + 15x^4-20x^5 .....etc
THEN integrate
(1/3)x^3 - (3/2)x^4 + 4x^5 -(10/3)x^6 ... etc
whoop, that is 3 x^5 not 4 x^5
Otherwise I am agreeing with Wolfram
Get it ?
To find the indefinite integral of the function ∫ x^2(1-x)^6 dx, you can use the method of integration by substitution. Here's a step-by-step explanation:
1. Let u = 1 - x. This choice of substitution is good because we have a factor of (1-x) raised to a power, which can simplify when using u.
2. Compute du/dx by differentiating both sides of the equation u = 1 - x with respect to x. This gives du/dx = -1.
3. Rearrange the equation to solve for dx in terms of du: dx = -du.
4. Substitute the expressions for x and dx in terms of u into the original integral, so it becomes: ∫ (1 - u)^2 * u^6 * (-du).
5. Simplify the expression by expanding the squared term: ∫ (1 - 2u + u^2) * u^6 * (-du).
6. Distribute the integrand to get three separate terms: ∫ (u^6 - 2u^7 + u^8) * (-du).
7. Distribute the negative sign and integrate term-by-term: - ∫ u^6 * du + 2 ∫ u^7 * du - ∫ u^8 * du.
8. Use the power rule for integration, which states that ∫ x^n dx = (x^(n+1))/(n+1) + C, where C is the constant of integration.
9. Applying the power rule to each term, the integral becomes: - (u^7)/7 + 2(u^8)/8 - (u^9)/9 + C.
10. Replace u with 1 - x: - ((1 - x)^7)/7 + 2((1 - x)^8)/8 - ((1 - x)^9)/9 + C.
Therefore, the indefinite integral of ∫ x^2(1-x)^6 dx is given by - ((1 - x)^7)/7 + 2((1 - x)^8)/8 - ((1 - x)^9)/9 + C, where C is the constant of integration.