Question

integration of x square -2x+1

Answer 1

Given,
integrant is x²- 2x+ 1.
we know that int if x^n with respect to x is {x^(n+1)}/n+1
see picture!!
.... Hope it helped you!!

Answer 2

$\displaystyle \sf \int ( {x}^{2} - 2x + 1)dx = \frac{ {x}^{3} }{3} - {x}^{2} + x + c$

Correct question :

$\displaystyle \sf \int ( {x}^{2} - 2x + 1)dx$

Given :

$\displaystyle \sf \int ( {x}^{2} - 2x + 1)dx$

To find :

Integrate the integral

Formula :

$\displaystyle \sf \int {x}^{n} dx = \frac{ {x}^{n + 1} }{n + 1} + constant$

Solution :

Step 1 of 2 :

Write down the given Integral

Here the given Integral is

$\displaystyle \sf \int ( {x}^{2} - 2x + 1)dx$

Step 2 of 2 :

Integrate the integral

$\displaystyle \sf \int ( {x}^{2} - 2x + 1)dx$

$\displaystyle \sf = \int {x}^{2} dx - \int 2x dx+ \int1dx$

$\displaystyle \sf = \int {x}^{2} dx -2 \int {x}^{1} dx+ \int {x}^{0} dx$

$\displaystyle \sf = \frac{ {x}^{2 + 1} }{2 + 1} - 2. \frac{ {x}^{1 + 1} }{1 + 1} + \frac{ {x}^{0 + 1} }{0 + 1} + c \: \: \: \: \bigg[ \: \because \int {x}^{n} dx = \frac{ {x}^{n + 1} }{n + 1} + constant\bigg]$

$\displaystyle \sf = \frac{ {x}^{3} }{3} - 2. \frac{ {x}^{2} }{2} + \frac{ {x}^{ 1} }{ 1} + c$

$\displaystyle \sf = \frac{ {x}^{3} }{3} - {x}^{2} + x + c$

Where c is integration constant

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