Final answer:
To approximate the definite integral using a power series, we can expand the function as a series and integrate each term. Then, we add up the individual integrals to find the approximate value.
Step-by-step explanation:
To approximate the definite integral, we can use a power series. In this case, we have the function f(x) = 1 + x^5. We can rewrite the function as a power series by expanding (1 + x)^5 using the Binomial theorem.
Following the steps of the Binomial theorem, expand (1 + x)^5:
- Identify the formula: (1 + x)^n = C(n, 0) * 1^n * x^0 + C(n, 1) * 1^(n-1) * x^1 + C(n, 2) * 1^(n-2) * x^2 + ... + C(n, n) * 1^0 * x^n
- For n = 5, the expansion is: (1 + x)^5 = C(5, 0) * 1^5 * x^0 + C(5, 1) * 1^4 * x^1 + C(5, 2) * 1^3 * x^2 + C(5, 3) * 1^2 * x^3 + C(5, 4) * 1^1 * x^4 + C(5, 5) * 1^0 * x^5
- Simplify the coefficients: (1 + x)^5 = 1 * x^0 + 5 * x^1 + 10 * x^2 + 10 * x^3 + 5 * x^4 + 1 * x^5
Now we can use this power series to compute the definite integral:
- Integrate each term of the power series: ∫[0,2] (1 + x^5) dx = ∫[0,2] (1*dx + 5*x*dx + 10*x^2*dx + 10*x^3*dx + 5*x^4*dx + x^5*dx)
- Solve each integral: ∫[0,2] (1*dx) = x | [0,2] = 2 - 0 = 2
- ∫[0,2] (x*dx) = (1/2)*x^2 | [0,2] = (1/2)*(2^2) - (1/2)*(0^2) = 2
- ∫[0,2] (x^2*dx) = (1/3)*x^3 | [0,2] = (1/3)*(2^3) - (1/3)*(0^3) = 8/3
- ∫[0,2] (x^3*dx) = (1/4)*x^4 | [0,2] = (1/4)*(2^4) - (1/4)*(0^4) = 4/4 = 1
- ∫[0,2] (x^4*dx) = (1/5)*x^5 | [0,2] = (1/5)*(2^5) - (1/5)*(0^5) = 32/5
- ∫[0,2] (x^5*dx) = (1/6)*x^6 | [0,2] = (1/6)*(2^6) - (1/6)*(0^6) = 64/6 = 32/3
Finally, add up the individual integrals: 2 + 2 + 8/3 + 1 + 32/5 + 32/3 ≈ 26.062667