Integral Calculus-1
Integral Calculus-1
Allen Star Batch
Grade 12
Question:
$I_1 = \int f(x) dx$ and $I_2 = \int_0^1 f(x) dx$ where $f(x) = x^2 \ln\left(1-x^2\right)$, then:
$I_1 = -\left\{\frac{x^3}{1.3} + \frac{x^5}{3.5} + \frac{x^7}{5.7} + \ldots\right\}$
$I_2 = \frac{2}{3}\ln 2 - \frac{8}{9}$
$I_1 = -\left\{\frac{x^5}{1.5} + \frac{x^7}{2.7} + \frac{x^9}{3.9} + \ldots\right\}$
$I_2 = \frac{2}{3}\ln 2 - \frac{5}{9}$
Step-by-Step Solution
Key Concept: Replace variable in series, multiply by $x^2$, integrate term-by-term, and use integration by parts with logarithmic functions.
Starting with the series $\ln(1-x) = -\left(x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \ldots\right)$, replace $x$ with $x^2$ to get $\ln(1-x^2) = -\left(x^2 + \frac{x^4}{2} + \frac{x^6}{3} + \frac{x^8}{4} + \ldots\right)$. Multiply by $x^2$ and integrate from 0 to 1 using integration by parts. The limit term vanishes as $x^3 - 1 \to 0$, yielding $\frac{1}{1.5} + \frac{1}{2.7} + \frac{1}{3.9} + \ldots = \frac{2}{3}\ln 2 - \frac{8}{9}$.
Correct Answer: 2,3