Definite Integration
Evaluation of definite integrals
Grade 12
Question:
<p>If \(I_n = \displaystyle\int_{-1}^{1} |x|\!\left(1 + x + \dfrac{x^2}{2} + \dfrac{x^3}{3} + \cdots + \dfrac{x^{2n}}{2n}\right)dx\), then:</p>
<p>(a) \(I_2 = \dfrac{4}{3}\)</p>
<p>(b) \(I_2 = \dfrac{7}{6}\)</p>
<p>(c) \(\displaystyle\lim_{n \to \infty} I_n = \dfrac{3}{2}\)</p>
<p>(d) \(\displaystyle\lim_{n \to \infty} I_n = \dfrac{5}{4}\)</p>
Step-by-Step Solution
Key Concept: Split the integral at x=0 using |x|=-x for x<0 and |x|=x for x>0, then exploit the even-odd properties of the integrand. The polynomial sum contains both even and odd powers, so only even-power terms survive after multiplication by |x|.
<p><strong>Step 1:</strong> Split the integral using the definition of |x|:</p><p>$$I_n = \int_{-1}^{0} (-x)\left(1 + x + \frac{x^2}{2} + \cdots + \frac{x^{2n}}{2n}\right)dx + \int_{0}^{1} x\left(1 + x + \frac{x^2}{2} + \cdots + \frac{x^{2n}}{2n}\right)dx$$</p><p><strong>Step 2:</strong> Separate even and odd power terms. When |x| multiplies odd powers (x, x³, ..., x^(2n-1)), the resulting integrand is odd, so these integrals over [-1,1] vanish.</p><p><strong>Step 3:</strong> Only even-power terms survive:</p><p>$$I_n = 2\int_{0}^{1} x\left(1 + \frac{x^2}{2} + \frac{x^4}{4} + \cdots + \frac{x^{2n}}{2n}\right)dx$$</p><p><strong>Step 4:</strong> Integrate term by term:</p><p>$$I_n = 2\int_{0}^{1} \left(x + \frac{x^3}{2} + \frac{x^5}{4} + \cdots + \frac{x^{2n+1}}{2n}\right)dx$$</p><p>$$= 2\left[\frac{x^2}{2} + \frac{x^4}{8} + \frac{x^6}{24} + \cdots + \frac{x^{2n+2}}{2n(2n+2)}\right]_0^1$$</p><p>$$= 2\left(\frac{1}{2} + \frac{1}{8} + \frac{1}{24} + \cdots + \frac{1}{2n(2n+2)}\right)$$</p><p>$$= 1 + \frac{1}{4} + \frac{1}{12} + \cdots + \frac{1}{n(2n+1)}$$</p><p>∴ Answer: BD</p>
Correct Answer: BD