Definite Integration
Comparison of Integrals
Grade 12
Question:
<p>If \(I_1 = \int_0^1 \frac{1+x^8}{1+x^4} dx\) and \(I_2 = \int_0^1 \frac{1+x^9}{1+x^3} dx\), then:</p>
<p>(a) \(I_1 > 1, I_2 < 1\)</p>
<p>(b) \(I_1 < 1, I_2 > 1\)</p>
<p>(c) \(1 < I_1 < I_2\)</p>
<p>(d) \(I_2 < I_1 < 1\)</p>
Step-by-Step Solution
Key Concept: Factor and simplify the rational integrands to make them easier to integrate, then evaluate and compare the resulting values.
<p>Simplify the integrands: $\frac{1+x^8}{1+x^4} = \frac{(1+x^4)^2 - 2x^4}{1+x^4} = (1+x^4) - \frac{2x^4}{1+x^4}$ and $\frac{1+x^9}{1+x^3} = \frac{(1+x^3)(1+x^6-x^3)}{1+x^3} = 1 + x^6 - x^3$. Evaluate both integrals over $[0,1]$ and compare. Both are less than 1, and $I_2 < I_1$.</p>
Correct Answer: D