Forany real number x, the value of ∫0x [x]dx is

Let k be an integer such that k≤x<k+1 then.∫0x [x]dx=∫0k [x]dx+∫kx [x]dx⇒∫0x [x]dx=∫01 [x]dx+∫12 [x]dx+….+∫k−1k [x]dx+∫kx [x]dx⇒∫0x [x]dx=∑r=1k ∫r−1r [x]dx+∫kx [x]dx⇒∫0x [x]dx=∑r=1k ∫r−1r (r−1)dx+∫kx kdx⇒∫0x [x]dx=∑r=1k (r−1)+k(x−k)⇒∫0x [x]dx=(1+2+…+k−1)+k(x−k)⇒∫0x [x]dx=k(k−1)2+kx−k2⇒∫0x [x]dx=kx−k(k+1)2=[x]x−[x]([x]+1)2

Forany real number x, the value of ∫0x [x]dx is

Let k be an integer such that k≤x<k+1 then.∫0x [x]dx=∫0k [x]dx+∫kx [x]dx⇒∫0x [x]dx=∫01 [x]dx+∫12 [x]dx+….+∫k−1k [x]dx+∫kx [x]dx⇒∫0x [x]dx=∑r=1k ∫r−1r [x]dx+∫kx [x]dx⇒∫0x [x]dx=∑r=1k ∫r−1r (r−1)dx+∫kx kdx⇒∫0x [x]dx=∑r=1k (r−1)+k(x−k)⇒∫0x [x]dx=(1+2+…+k−1)+k(x−k)⇒∫0x [x]dx=k(k−1)2+kx−k2⇒∫0x [x]dx=kx−k(k+1)2=[x]x−[x]([x]+1)2

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

Forany real number x, the value of $\int_{0}^{x} [x] d x$ is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$x [x] - [x] ([x] + 1)$

none of these

$x [x]$

$x [x] - \frac{1}{2} [x] ([x] + 1)$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Let k be an integer such that $k \leq x < k + 1$ then.

$\begin{array}{l} \int_{0}^{x} [x] d x = \int_{0}^{k} [x] d x + \int_{k}^{x} [x] d x \\ \Rightarrow \int_{0}^{x} [x] d x = \int_{0}^{1} [x] d x + \int_{1}^{2} [x] d x + \dots . + \int_{k - 1}^{k} [x] d x + \int_{k}^{x} [x] d x \\ \Rightarrow \int_{0}^{x} [x] d x = \sum_{r = 1}^{k} \int_{r - 1}^{r} [x] d x + \int_{k}^{x} [x] d x \\ \Rightarrow \int_{0}^{x} [x] d x = \sum_{r = 1}^{k} \int_{r - 1}^{r} (r - 1) d x + \int_{k}^{x} k d x \\ \Rightarrow \int_{0}^{x} [x] d x = \sum_{r = 1}^{k} (r - 1) + k (x - k) \\ \Rightarrow \int_{0}^{x} [x] d x = (1 + 2 + \dots + k - 1) + k (x - k) \end{array}$

$\begin{array}{l} \Rightarrow \int_{0}^{x} [x] d x = \frac{k (k - 1)}{2} + k x - k^{2} \\ \Rightarrow \int_{0}^{x} [x] d x = k x - \frac{k (k + 1)}{2} = [x] x - \frac{[x] ([x] + 1)}{2} \end{array}$