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Mathematics - Definite Integration Question with Solution | TestHub

MathematicsDefinite IntegrationProperties of definite integrationHard2 minPYQ_2020

MathematicsHardnumerical

Let $[t]$ denote the greatest integer less than or equal to $t$ . Then the value of $\int_{1}^{2} |2 x - [3 x]| d x$ is

Answer:

1.00

Solution:

$\int_{1}^{2} |3 x - [3 x] - x| d x$

$= \int_{1}^{2} |\{3 x\} - x| d x$

$= \int_{1}^{2} (x - \{3 x\}) d x$

$= \int_{1}^{2} x d x - \int_{1}^{2} \{3 x\} d x$

$= {(\frac{x^{2}}{2})}_{1}^{2} - 3 \int_{0}^{1 / 3} 3 x d x$

$= (\frac{4}{2} - \frac{1}{2}) - 9 {(\frac{x^{2}}{2})}_{0}^{1 / 3}$

$= \frac{3}{2} - \frac{9}{2} ((1 / 3)^{2} - 0^{2})$

$= 1$

Stream:JEESubject:MathematicsTopic:Definite IntegrationSubtopic:Properties of definite integration

⏱ 2mℹ️ Source: PYQ_2020

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