use the fourth taylor polynomial of
f(x)=ln(1+x) to approximate
∫0.2 ln (x+1)/x dx
0.1
To approximate the integral ∫0.2 ln(x+1)/x dx using the 4th Taylor polynomial of f(x) = ln(1+x), we can follow these steps:
Step 1: Write the Taylor polynomial of f(x) = ln(1+x) up to the 4th degree.
The general formula for the nth degree Taylor polynomial of f(x) centered at x = a is:
Pₙ(x) = f(a) + f'(a)(x-a) + (f''(a)/2!)(x-a)² + ... + (fⁿ(a)/𝒩!)(x-a)ⁿ
The 4th degree Taylor polynomial of f(x) = ln(1+x) centered at a = 0 is given by:
P₄(x) = f(0) + f'(0)x + (f''(0)/2!)x² + (f'''(0)/3!)x³ + (f⁴(0)/4!)x⁴
Step 2: Calculate the coefficients of the polynomial.
To find the coefficients of the polynomial, we need to find the values of f(0), f'(0), f''(0), f'''(0), and f⁴(0):
f(0) = ln(1+0) = ln(1) = 0
f'(x) = 1/(1+x)
f'(0) = 1/(1+0) = 1
f''(x) = -1/(1+x)²
f''(0) = -1/(1+0)² = -1
f'''(x) = 2/(1+x)³
f'''(0) = 2/(1+0)³ = 2
f⁴(x) = -6/(1+x)⁴
f⁴(0) = -6/(1+0)⁴ = -6
Therefore, the coefficients of the 4th degree Taylor polynomial are:
P₄(x) = 0 + 1x + (-1/2!)x² + (2/3!)x³ + (-6/4!)x⁴
Step 3: Evaluate the integral using the Taylor polynomial.
Now we can use the Taylor polynomial to approximate the integral:
∫₀.₂ ln(x+1)/x dx ≈ ∫₀.₂ P₄(x) dx
Since the given integral limits are from 0 to 0.2, we can substitute x with 0.2 in the polynomial function:
∫₀.₂ ln(x+1)/x dx ≈ ∫₀.₂ [0 + 1x + (-1/2!)x² + (2/3!)x³ + (-6/4!)x⁴] dx
∫₀.₂ ln(x+1)/x dx ≈ 0 + (1/2)x² - (1/6)x³ + (-6/120)x⁴
Step 4: Evaluate the approximate integral.
Now we can plug in the limits of integration (0 and 0.2) into the polynomial and calculate the integral:
∫₀.₂ ln(x+1)/x dx ≈ [1/2 * (0.2)²] - [1/6 * (0.2)³] + [-6/120 * (0.2)⁴]
Simplifying the expression, we get:
∫₀.₂ ln(x+1)/x dx ≈ 0.0204
Therefore, the approximate value of the integral ∫₀.₂ ln(x+1)/x dx, using the 4th degree Taylor polynomial of f(x) = ln(1+x), is approximately 0.0204.