Definite Integration
General
Grade 12
Question:
The value of $\int_{1}^{2} (x^{[x^2]} + [x^2]^x) dx$, where $[.]$ denotes the greatest integer function, is equal to -
$\frac{5}{4} + \sqrt{3} + (2^{\sqrt{3}} - 2^{\sqrt{2}}) + \frac{1}{\log 3} (9 - 3^{\sqrt{3}})$
$\frac{5}{4} + \sqrt{3} + \frac{\sqrt{2}}{3} + \frac{1}{\log 2} (2^{\sqrt{3}} - 2^{\sqrt{2}}) + \frac{1}{\log 3} (9 - 3^{\sqrt{3}})$
$\frac{5}{4} + \frac{\sqrt{2}}{3} + \frac{1}{\log 2} (2^{\sqrt{3}} - 2^{\sqrt{2}}) + \frac{1}{\log 3} (9 - 3^{\sqrt{3}})$
none of these
Step-by-Step Solution
Key Concept: General
We have, $I = \int_{1}^{2} (x^{[x^2]} + [x^2]^x) dx = \int_{1}^{\sqrt{2}} (x + 1) dx + \int_{\sqrt{2}}^{\sqrt{3}} (x^2 + 2^x) dx + \int_{\sqrt{3}}^{2} (x^3 + 3^x) dx$<br/>$= \left( \frac{x^2}{2} + x \right)_{1}^{\sqrt{2}} + \left( \frac{x^3}{3} + \frac{2^x}{\log 2} \right)_{\sqrt{2}}^{\sqrt{3}} + \left( \frac{x^4}{4} + \frac{3^x}{\log 3} \right)_{\sqrt{3}}^{2}$<br/>$= \frac{5}{4} + \sqrt{3} + \frac{\sqrt{2}}{3} + \frac{1}{\log 2} (2^{\sqrt{3}} - 2^{\sqrt{2}}) + \frac{1}{\log 3} (3^2 - 3^{\sqrt{3}})$
Correct Answer: B