Question

If [α]  denotes the greatest integer less than or equal to a for α  € R, then the value of the integral \(\mathop \smallint \nolimits_0^{1.7} \left[ {{x^2}} \right]dx\) is equal to

1. 2.4 + √ 2
2. 2.4 - √ 2
3. \(2.4 + \frac{1}{\sqrt2}\)
4. \(2.4 - \frac{1}{\sqrt2}\)

Answer 1

Correct Answer - Option 2 : 2.4 - √ 2

Calculation:

I = \(\rm \mathop \smallint \nolimits_0^{1.7} \left[ {{x^2}} \right]dx\)

I = \(\rm \mathop \smallint \nolimits_0^{1} \left[ {{x^2}} \right]dx\) + \(\rm \mathop \smallint \nolimits_1^{√2} \left[ {{x^2}} \right]dx\) + \(\rm \mathop \smallint \nolimits_{√2}^{1.7} \left[ {{x^2}} \right]dx\)

∵ For 0 ≤ x < 1, [x²] = 0

For 1 ≤ x < √2, [x²] = 1

For √2 ≤ x < √3(1.732), [x²] = 2

∴ I = \(\rm \mathop \smallint \nolimits_0^{1} \left[ {{0}} \right]dx\) + \(\rm \mathop \smallint \nolimits_1^{√2} \left[ {{1}} \right]dx\) + \(\rm \mathop \smallint \nolimits_{√2}^{1.7} \left[ {{2}} \right]dx\)

I = \(\rm \mathop \smallint \nolimits_1^{√2} dx\) + \(\rm2 \mathop \smallint \nolimits_{√2}^{1.7} dx\)

I = \(\rm \left[ {{x}} \right]_1^{√2}\) + \(\rm 2 \left[ {{x}} \right]_{√2}^{1.7}\)

I = \(√2-1+2(1.7-√2)\)

I = 2.4 - √2