Question

$\int\limits^2_0 (6x²-2x+7)dx ,Obtain the given integrals as the limit of a sum.$

Answer 1

HELLO DEAR,


GIVEN:-
$\sf{\int\limits^2_0 (6x^2-2x+7)dx}$

so, $\sf{\int\limits^2_0 (6x^2*dx - 2x*dx + 7*dx)}$

$\sf{ [6x^3/3 - 2x^2/2 + 7x]^2_0}$

$\sf{[2*(8) - (4) + 7(2)]}$

$\sf{[16 - 4 + 14]}$

$\sf{26}$


I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

we have to get the value of $\int\limits^2_0{(6x^2-2x+7)}\,dx$

we know, $\int{x^n}\,dx=\left[\frac{x^{n+1}}{n+1}\right]+C$

$\text{here,}\int\limits^2_0{(6x^2-2x+7)}\,dx\\\\\\=\int\limits^2_0{6x^2}\,dx-\int\limits^2_0{2x}\,dx+7\int\limits^2_0{dx}\\\\\\=6\int\limits^2_0{x^2}\,dx-2\int\limits^2_0{x}\,dx+7\int\limits^2_0{dx}\\\\\\=6\left[\frac{x^3}{3}\right]^2_0-2\left[\frac{x^2}{2}\right]^2_0+7\left[x\right]^2_0\\\\\\=2\times(2)^3-(2)^2+7\times2\\\\\\=16-4+14=26$