Question

$\int\limits^4_2(2x-1)dx ,Obtain the given integrals as the limit of a sum.$

Answer 1

we have to find the value of $\int\limits^4_2{(2x-1)}\,dx$

$\int\limits^4_2{(2x-1)}\,dx$

= $\int\limits^4_2{2x}\,dx-\int\limits^4_2{dx}$

= $\left[\begin{array}{c}2\frac{x^2}{2}\end{array}\right]^4_2-[x]^4_2$

=$[x^2]^4_2-[x]^4_2$

= $(4^2-2^2)-(4-2)$

= $(16-4)-(2)$

= $10$

Answer 2

HELLO DEAR,

GIVEN:-
$\int\limits^4_2{(2x - 1)}\,dx$

$\int\limits^4_2\{{2x.dx - dx}\}$

$\bold{[2x^2/2 - x]^4_2}$

$\bold{[(4)^2 - (2)^2 - \{4 - 2\}]}$

$\bold{[16 - 4 - 2]}$

$\bold{\Rightarrow 10}$

I HOPE ITS HELP YOU DEAR,
THANKS