MHT CET · Maths · Definite Integration
\(\int_0^4 x[x] d x=(\) where \([x]\) denotes greatest integer function not greater than \(\mathrm{x}\) )
- A 17
- B 24
- C \(\frac{21}{2}\)
- D \(\frac{33}{2}\)
Answer & Solution
Correct Answer
(A) 17
Step-by-step Solution
Detailed explanation
Let \(I=\int_0^4 x[x] d x\)
\( \therefore I=\int_0^1(0) d x+\int_1^2 x d x+\int_2^3 2 x d x+\) \(\int_3^4 3 x d x \)
\( =\left[\frac{x^2}{2}\right]_1^2+\left[\frac{2 x^2}{2}\right]_2^3+\left[\frac{3 x^2}{2}\right]_3^4 \)
\( =\frac{1}{2}(4-1)+(9-4)+\left(\frac{3}{2}\right)(16-9)=\) \(\frac{3}{2}+5+\frac{21}{2}=17\)
\( \therefore I=\int_0^1(0) d x+\int_1^2 x d x+\int_2^3 2 x d x+\) \(\int_3^4 3 x d x \)
\( =\left[\frac{x^2}{2}\right]_1^2+\left[\frac{2 x^2}{2}\right]_2^3+\left[\frac{3 x^2}{2}\right]_3^4 \)
\( =\frac{1}{2}(4-1)+(9-4)+\left(\frac{3}{2}\right)(16-9)=\) \(\frac{3}{2}+5+\frac{21}{2}=17\)
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