Let [t] denote the largest integer less than or equal to t . If ∫_0^3([x^2]+[(x^2)/(2)]) d x=a+b 2 - 3 - 5 +c 6 - 7 , where a, b, c ∈ Z , then a+b+c is equal to_______

Correct answer: 0 aligned & ∫_0^3[x^2] d x+∫_0^3[(x^2)/(2)] d x & =∫_0^1 0 d x+∫_1^ 12 1 d x+∫_ 2 ^ 3 2 d x aligned aligned & +∫_ 3 ^2 3 dx+∫_2^ 5 4 dx+∫_ 5 ^ 6 5 dx & +∫_ 6 ^ 7 6 dx+∫_ 7 ^ 8 7 dx+∫_ 8 ^3 8 dx & +∫_0^ 2 0 dx+∫_ 2 ^2 1 dx & +∫_2^ 6 2 dx+∫_ 6 ^ 8 3 dx+∫_ 8 ^3 4 dx=31-6 2 - 3 - 5 & -2 6 - 7 & a=31 ~b=-6 c=-2 & a+b+c=31-6-2=23 aligned

JEE Main 20246 AprilEvening Shift

Exam: JEE Main
Year: 2024
Date: 6 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Definite Integration

JEE Main 2024 · 6 April · Evening Shift · Mathematics · Definite Integration

Let [t] denote the largest integer less than or equal to t . If...

Let

[t]

denote the largest integer less than or equal to

t

. If

\int_0^3\left(\left[x^2\right]+\left[\frac{x^2}{2}\right]\right) \mathrm{d} x=\mathrm{a}+\mathrm{b} \sqrt{2}-\sqrt{3}-\sqrt{5}+\mathrm{c} \sqrt{6}-\sqrt{7}

, where

\mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathbf{Z}

, then

\mathrm{a}+\mathrm{b}+\mathrm{c}

is equal to_______

Correct answer

Step-by-step solution

\begin{aligned} & \int_0^3\left[x^2\right] d x+\int_0^3\left[\frac{x^2}{2}\right] d x \\ & =\int_0^1 0 d x+\int_1^{12} 1 d x+\int_{\sqrt{2}}^{\sqrt{3}} 2 d x\end{aligned}

\begin{aligned} & +\int_{\sqrt{3}}^2 3 \mathrm{dx}+\int_2^{\sqrt{5}} 4 \mathrm{dx}+\int_{\sqrt{5}}^{\sqrt{6}} 5 \mathrm{dx} \\ & +\int_{\sqrt{6}}^{\sqrt{7}} 6 \mathrm{dx}+\int_{\sqrt{7}}^{\sqrt{8}} 7 \mathrm{dx}+\int_{\sqrt{8}}^3 8 \mathrm{dx} \\ & +\int_0^{\sqrt{2}} 0 \mathrm{dx}+\int_{\sqrt{2}}^2 1 \mathrm{dx} \\ & +\int_2^{\sqrt{6}} 2 \mathrm{dx}+\int_{\sqrt{6}}^{\sqrt{8}} 3 \mathrm{dx}+\int_{\sqrt{8}}^3 4 \mathrm{dx}=31-6 \sqrt{2}-\sqrt{3}-\sqrt{5} \\ & -2 \sqrt{6}-\sqrt{7} \\ & \mathrm{a}=31 \quad \mathrm{~b}=-6 \quad \mathrm{c}=-2 \\ & \mathrm{a}+\mathrm{b}+\mathrm{c}=31-6-2=23\end{aligned}

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