$\int_{0}^{6} x^{\{x\}} d x $ where {} in the G.I.F

Question

An integral question from the NSEA-2018(question 70 in this solutions booklet) Paper.

Q.Find the integer closest to the integral $$ \int_{0}^{6} x^{\{\sqrt{x}\}} d x $$ , where {x} denotes the largest integer not exceeding x.* Corrected a latex error here *

The question is filled with strangeness as they've used mod(|x|) in the Integral and not [x]but have described {x} and not [x]as the G.I.F.) * Corrected a mistake originally added here-ignore this *. My thoughts:- 1. An indefinite integral of $x^{x}$ is unknown. But not sure about definite.

2. I tried substitutions and King's property but failed to reach anywhere meaningful.

3. Unsure on how to tackle a fractional part in the power.

Answer 1

This question is crippled with mistakes.

If first shows the exponent $|\sqrt x|$, where the absolute value is useless. Then if defines $\{x\}$ though this is not used, and assigns it the meaning of the floor function !?

Finally the solution is given for

$$\int_0^6 x^{\lfloor\sqrt x\rfloor}dx$$ by decomposing the range, and the computation is wrong (but the final answer is right)

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In your question, you dropped the square root and worsened the ambiguity.

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$$\int_1^6 x^{\lfloor\sqrt x\rfloor}dx=\int_0^1dx+\int_1^4x\,dx+\int_4^6x^2\,dx=1+\frac{4^2-1}2+\frac{6^3-4^2}3=59.16\cdots.$$

Answer 2

Here $[.]$ is GIF. $$\int_{0}^{6} x^{[x]} dx==\int_{0}^{1} x^0 dx+ \int_{1}^{2} x^1 dx+\int_{2}^{3} x^2 dx+\int_{3}^{4} x^3 dx.+\int_{4}^{5} x^4 dx....+\int_{5}^{6} x^5 dx=\frac{338677}{60} $$