Let $\int\limits_{ - 3}^t {\left\lfloor x \right\rfloor } dx = 51.75$. What is $t$?

Question

Let $\int\limits_{ - 3}^t {\left\lfloor x \right\rfloor } dx = 51.75$.

What is $t$?

Answer 1

You can break up the integrals into smaller parts. Start like this: $$\int_{-3}^t {\left\lfloor x \right\rfloor }~dx=\int_{-3}^{-2} -3~dx+\int_{-2}^{-1}-2~dx+\int_{-1}^{0}-1~dx+\cdots+\int_a^ta~dx \tag{1}=51.75$$ Where $a$ is some positive integer you must find. Since the integrands are constant, all you need to do is integrate repeatedly until you get close to $51.75$.

Notice that all you need to do is find the area of rectangles of width $1$:

So, for the area from $x=-3$ to $x=0$, we have: $$\int_{-3}^0 {\left\lfloor x \right\rfloor }~dx=(-3)+(-2)+(-1)=-6$$ Therefore, the area from $x=0$ to $x=t$ must be equal to: $$\int_{0}^t {\left\lfloor x \right\rfloor }~dx=51.75+6=57.75$$ Therefore: $$\int_{0}^t {\left\lfloor x \right\rfloor }~dx=\int_{0}^{1} 0~dx+\int_{1}^{2}1~dx+\int_{2}^{3}2~dx+\cdots+\int_a^t a~dx=57.75 \tag{2}$$

Can you continue? Hint: Triangular Numbers.

Answer 2

Well

$\int_{n}^{n+1} \lfloor x \rfloor dx = \int_{n}^{n+1} n dx = n$, for $n \in \mathbb Z$, obviously.

So $\int_{-3}^t \lfloor x \rfloor dx = \sum_{n = -3}^{\lfloor t-1 \rfloor}\int_n^{n+1} \lfloor x \rfloor dx + \int_{\lfloor t \rfloor}^t \lfloor x \rfloor dx=$

$\sum_{n = -3}^{\lfloor t-1 \rfloor} n + [t]*(t-\lfloor t \rfloor)=$

$-6 + \frac {(t-1)t}2 + [t]*\{t\}=51.75$ where $\{t\}$ is the fractional part of $t$, i.e. $t - \lfloor t \rfloor$.

So $\frac{([t]-1)t}2= 1 + 2 + 3+ ..... + ([t]- 1) \le 57$. That is $[t]-1 = 10$ as $\sum_{k=1}^{10} k = \frac {10*11}2 =55$.

So $[t] = 11$, (i.e. $11 \le t < 12$) and $11(t - 11) = 2.75=\frac {11}4$ so $t-11 = \frac 14$ so $t = 11\frac 14$.

Answer 3

Hint : For these type of questions, always let $t= I + f$ where $I \in \mathbb Z$ and $0 \le f<1$.

And put $\lfloor x \rfloor = x- \lbrace x \rbrace $

Where $\lbrace . \rbrace$ denotes fractional part of $x$

Now, use periodicity of $\lbrace x \rbrace$ and separate integral into two parts i.e. $-3$ to $I$ and $I$ to $I +f$.

Solution : As said above -

$\int\limits_{ - 3}^t {\left\lfloor x \right\rfloor } dx = \int\limits_{ - 3}^{I+f} {x-\lbrace x \rbrace } dx $

$=\int\limits_{ - 3}^{I+f} {x} dx - \int\limits_{ - 3}^{I+f} {\lbrace x \rbrace } dx $

$=\int\limits_{ - 3}^{I+f} {x} dx - \int\limits_{ - 3}^{I} {\lbrace x \rbrace } dx - \int\limits_{ I}^{I+f} {\lbrace x \rbrace } dx$

$= \frac { x^{ 2 } }{ 2 } | _{ -3 }^{ I+f } -(I+3)\int\limits_{ 0}^{1} {\lbrace x \rbrace } dx-\int\limits_{ 0}^{f} {\lbrace x \rbrace } dx $ Since $\lbrace x \rbrace $ is periodic with period $1$.

For $0$ to $1$ , $\lbrace x \rbrace=x$

$= \frac {(I+f)^2}{2}-\frac {(-3)^2}{2}- (I+3)\times \frac{1}{2}-\frac{f^2}{2} $

$(I+f)^2-9- (I+3)-f^2 = 103.5$

$I^2+2If-I=115.5 $ ............ $(\star)$

$I^2+I \ge 116$ and $I^2-I \le 115$. Since $0 \le f <1$

$\rightarrow I= 11$

Put $I=11$ in $(\star)$, you will get $f=0.25$

Hence, $t=I+f=11.25$