Question

integration of mod x from 0 to 1

Answer 1

EXPLANATION.

$\sf \implies \int\limits^1_0 {|x|} \, dx$

As we know that,

$\sf \implies \int\limits^b_a {f(x)} \, dx = \int\limits^c_a {f(x)} \, dx + \int\limits^b_c {f(x)} \, dx \ \ where \ \ a < c < b$

Integral is broken at points of discontinuity or at the points where definition is changes.

Using the formula in equation, we get.

$\sf \implies \int\limits^\frac{1}{2} _0 {(-x)} \, dx \ + \int\limits^1_\frac{1}{2} {(x)} \, dx$

$\sf \implies - \bigg[\dfrac{x^{2} }{2} \bigg]_0^\frac{1}{2} + \bigg[\dfrac{x^{2} }{2} \bigg]_\frac{1}{2}^1$

Put the limits in the equation, we get.

Put upper limit first then put the lower limit in equation, we get.

$\sf \implies -\bigg[\dfrac{\bigg(\dfrac{1}{2} \bigg)^{2} }{2} - \dfrac{(0)^{2} }{2} \bigg] + \bigg[\dfrac{(1)^{2} }{2} - \dfrac{\bigg(\dfrac{1}{2} \bigg)^{2} }{2} \bigg]$

$\sf \implies -\bigg[\dfrac{\bigg(\dfrac{1}{4}\bigg) }{2} \bigg] + \bigg[\dfrac{1}{2} - \dfrac{\bigg(\dfrac{1}{4}\bigg) }{2} \bigg]$

$\sf \implies -\bigg[\dfrac{1}{8} \bigg] + \bigg[\dfrac{1}{2} - \dfrac{1}{8} \bigg]$

$\sf \implies -\bigg[\dfrac{1}{8} \bigg] + \bigg[\dfrac{4 - 1}{8} \bigg]$

$\sf \implies -\bigg[\dfrac{1}{8} \bigg] + \bigg[\dfrac{3}{8} \bigg]$

$\sf \implies \bigg[\dfrac{2}{8} \bigg] = \dfrac{1}{4} = answer$

                                                                                                                   

MORE INFORMATION.

Some important formulae.

$\sf \implies \int\limits^\frac{\pi}{2} _0 {log sin(x)} \, dx = \int\limits^\frac{\pi}{2} _0 {log cos(x)} \, dx= -\bigg(\dfrac{\pi}{2} \bigg)log(2)$

$\sf \implies \int\limits^\frac{\pi}{2} _0 { sin^{n} (x)} \, dx = \int\limits^\frac{\pi}{2} _0 {cos^{n} (x)} \, dx = \bigg(\dfrac{n - 1}{n} \bigg) \bigg(\dfrac{n - 3}{n - 2} \bigg) ..\dfrac{2}{3} . 1 ( n \ is \ odd)$

$\sf \implies \int\limits^\frac{\pi}{2} _0 { sin^{n} (x)} \, dx = \int\limits^\frac{\pi}{2} _0 { cos^{n} (x)} \, dx = \bigg(\dfrac{n - 1}{n} \bigg) \bigg(\dfrac{n - 3}{n - 2} \bigg) ..\dfrac{1}{2} \times \dfrac{\pi}{2} ( n \ is \ even )$

$\sf \implies \int\limits^\frac{\pi}{2} _0 {sin^{m} (x)cos^{n} (x)} \, dx = \dfrac{(m - 1)(m - 3)..(2 \ or \ 1)( n - 1)(n - 3)..(2 \ or \ 1)}{(m + n)(m + n - 2)..( 2 \ or \ 1)} \times (1 \ or \ \dfrac{\pi}{2} )$

Answer 2

Answer:

$\sf \implies \int\limits^1_0 {|x|} \, dx$

As we know that,

$\sf \implies \int\limits^b_a {f(x)} \, dx = \int\limits^c_a {f(x)} \, dx + \int\limits^b_c {f(x)} \, dx \ \ where \ \ a < c < b$

Integral is broken at points of discontinuity or at the points where definition is changes.

Using the formula in equation, we get.

$\sf \implies \int\limits^\frac{1}{2} _0 {(-x)} \, dx \ + \int\limits^1_\frac{1}{2} {(x)} \, dx$

$\sf \implies - \bigg[\dfrac{x^{2} }{2} \bigg]_0^\frac{1}{2} + \bigg[\dfrac{x^{2} }{2} \bigg]_\frac{1}{2}^1$

Put the limits in the equation, we get.

Put upper limit first then put the lower limit in equation, we get.

$\sf \implies -\bigg[\dfrac{\bigg(\dfrac{1}{2} \bigg)^{2} }{2} - \dfrac{(0)^{2} }{2} \bigg] + \bigg[\dfrac{(1)^{2} }{2} - \dfrac{\bigg(\dfrac{1}{2} \bigg)^{2} }{2} \bigg]$

$\sf \implies -\bigg[\dfrac{\bigg(\dfrac{1}{4}\bigg) }{2} \bigg] + \bigg[\dfrac{1}{2} - \dfrac{\bigg(\dfrac{1}{4}\bigg) }{2} \bigg]$

$\sf \implies -\bigg[\dfrac{1}{8} \bigg] + \bigg[\dfrac{1}{2} - \dfrac{1}{8} \bigg]$

$\sf \implies -\bigg[\dfrac{1}{8} \bigg] + \bigg[\dfrac{4 - 1}{8} \bigg]$

$\sf \implies -\bigg[\dfrac{1}{8} \bigg] + \bigg[\dfrac{3}{8} \bigg]$

$\sf \implies \bigg[\dfrac{2}{8} \bigg] = \dfrac{1}{4}$