Question #0c324

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Answer 1 · 2017-02-05T07:36:33

$-10/3$ .

Comparing the given integral with $int_b^a f(x)dx," we have, "b=2, a=0, f(x)=3-x^2$ .

Hence, $Deltax=(a-b)/n=(0-2)/n=-2/n=h," say, ":. nto oo, h to 0$ .

$"Also, "x_i=b+iDeltax=2+i(-2/n)=2+ih.$

Note that, $nh=-2.$

$:. I=int_2^0 (3-x^2)dx$

$=lim_(n to oo) sum_(i=1)^(i=n)f(x_i)Deltax$

Since, $f(x_i)=f(2+ih)=3-(2+ih)^2=-1-4ih-i^2h^2,$

$I=lim_(n to oo,htoo) sum_(i=1)^(i=n)(-1-4ih-i^2h^2)h$

$=limsum(-h-4ih^2-i^2h^3)$

$=lim[-sumh-4h^2sumi-h^3sumi^2]$

$=lim[-hsum1-(4h^2){(n)(n+1)}/2-(h^3){(n)(n+1)(2n+1)}/6]$

$=lim_(hto0)[-hn-2(hn)(hn+h)-1/6(hn)(hn+h)(2hn+h)]$

$=[-(-2)-2(-2)(-2+0)-1/6(-2)(-2+0){2(-2)+0}]......[because, nh=-2]$

$=2-8+8/3=-6+8/3$

$:. I=-10/3.$

Verification :-

$int_2^0 (3-x^2)dx=[3x-x^3/3]_2^0=0-(6-8/3)=-10/3.$

Enjoy Maths!