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Definite Integration question

2010 · Shift 1 · Q49
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Definite Integration question

2010 · Shift 1 · Q49

JEE AdvancedMathematicsDefinite IntegrationMCQ+4 / −1
The value of lim⁡x→01x3∫0xtln⁡(1+t)t4+4dt\mathop {\lim }\limits_{x \to 0} {1 \over {{x^3}}}\int\limits_0^x {{{t\ln \left( {1 + t} \right)} \over {{t^4} + 4}}} dtx→0lim​x31​0∫x​t4+4tln(1+t)​dt is
  1. A
    000
  2. B
    112{1 \over 12}121​
  3. C
    124{1 \over 24}241​
  4. D
    164{1 \over 64}641​
View written solutionFree

Correct answer: B

  1. We need to evaluate L=lim⁡x→01x3∫0xtln⁡(1+t)t4+4 dt.L=\lim_{x\to 0}\frac{1}{x^3}\int_0^x \frac{t\ln(1+t)}{t^4+4}\,dt.L=limx→0​x31​∫0x​t4+4tln(1+t)​dt.

  2. Let f(t)=tln⁡(1+t)t4+4.f(t)=\frac{t\ln(1+t)}{t^4+4}.f(t)=t4+4tln(1+t)​. As t→0t\to 0t→0, expand the numerator and denominator near 000.

We know ln⁡(1+t)=t−t22+t33+O(t4).\ln(1+t)=t-\frac{t^2}{2}+\frac{t^3}{3}+O(t^4).ln(1+t)=t−2t2​+3t3​+O(t4). Hence tln⁡(1+t)=t(t−t22+t33+O(t4))=t2−t32+t43+O(t5).t\ln(1+t)=t\left(t-\frac{t^2}{2}+\frac{t^3}{3}+O(t^4)\right)=t^2-\frac{t^3}{2}+\frac{t^4}{3}+O(t^5).tln(1+t)=t(t−2t2​+3t3​+O(t4))=t2−2t3​+3t4​+O(t5). Also, t4+4=4(1+t44),t^4+4=4\left(1+\frac{t^4}{4}\right),t4+4=4(1+4t4​), so 1t4+4=14(1+O(t4)).\frac{1}{t^4+4}=\frac{1}{4}\left(1+O(t^4)\right).t4+41​=41​(1+O(t4)). Therefore,

=\frac{t^2}{4}+O(t^3).$$ 3. Now integrate from $0$ to $x$: $$\int_0^x f(t)\,dt=\int_0^x \left(\frac{t^2}{4}+O(t^3)\right)dt =\frac{1}{4}\cdot \frac{x^3}{3}+O(x^4) =\frac{x^3}{12}+O(x^4).$$ 4. Divide by $x^3$: $$\frac{1}{x^3}\int_0^x f(t)\,dt =\frac{1}{x^3}\left(\frac{x^3}{12}+O(x^4)\right) =\frac{1}{12}+O(x).$$ Thus, $$L=\frac{1}{12}.$$ 5. Checking options: - A: $0$ ❌ - B: $\frac{1}{12}$ ✅ - C: $\frac{1}{24}$ ❌ - D: $\frac{1}{64}$ ❌ So the correct option is **B**.
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