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!