InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
योगफल की सिमा के रूप में `int_(1)^(4)(x^2-1)dx` का मान ज्ञात कीजिएः। |
|
Answer» यहाँ a=1, b=4 `f(x)=x^(2)-x, h=(b-a)/n=3/n "या" nh=3` योगफल की सिमा की परिभाषा से, `underset(a)overset(b)intf(x)dx=underset(h to 0)lim h[f(a)+f(a+h)+f(a+2h)+....+f{a+(n-1)h}]` `underset(1)overset(4)int f(x^(2)-x)dx=underset(h to 0)lim h[f(1)+f(1+h)+f(1+2h)+....+f{1+(n-1)h}]........(1)` अब `f(1)=(1)^(2)-1=0` `f(1+h)=(1+h^(2))-(1+h)` `=(1+2h+h^(2))-1-h=h^(2)+h` `f(1+2h)=(1+2h)^(2)-(1+2h)` `=1+4h+4h^(2)-1-2h=4h^(2)+2h` `f[1+(n-1)h]` `=[1+(n-1)h]^(2)-[1+(n-1)h]` `=1+2(n-1)h+(n-1)^(2)h^(2)-1-(n-1)h` `=(n-1)^(2)h^(2)+(n-1)h` उपरोक्त मनो को समी (1) में रखने पर `underset(1)overset(4)int(x^(2)-x)dx` `=underset(h to 0)lim h[0+(h^(2)+h)+(4h^(2)+2h)+.....+{(n-1)^(2)h^(2)+(n-1)h}]` `=underset(h to 0)lim h[h^(2){1^2+2^2+.....+(n-1)^2}+h{1+2+...+(n-1)}]` `=underset(h to 0)lim [h^2 xx(n(n-1))/6+hxx(n(n-1))/(2)]` `=underset(h to 0)lim [(nh(nh-h)(2nh-1))/(6)+(nh(nh-h))/(2)]` `=underset(h to 0)lim [(3(3-h)(6-h))/6+(3(3-h))/(2)]` `=(3(3-0)(6-0))/6+(3(3-0))/(2)=9+9/2=27` |
|