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Consider the function defined implicity by the equation y^(2)-2ye^(sin^(-1)x)+x^(2)-1+[x]+e^(2sin ^(-1)x)=0("where [x] denotes the greatest integer function"). The area of the region bounded by the curve and the line x=-1 is |
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Answer» `PI+1` sq. units `(y-e^(sin^(-1)X))^(2)=2-x^(2)` `y=e^(sin-1x)pmsqrt(2-x^(2))` `A=overset(0)underset(-1)int(e^(sin^(-1)x)+sqrt(2-x^(2)))-(e^(sin^(-1)x)-sqrt(2-x^(2)))dx` `=2overset(0)underset(-1)intsqrt(2-x^(2))dx` `=2((1)/(2)xsqrt(2-x^(2)) :|_(-1)^(0)+(2)/(2)sin^(-1)""(x)/(sqrt(2)):|_(-1)^(0))` `=[1+2(0-(-(pi)/(54)))]` `=(pi)/(2)+1` sq. units. `"For "0lexlt1,y=sin^(-1)xpmsqrt(1-x^(2))` `A=2overset(1)underset(0)intsqrt(1-x^(2))dx` `=2[(x)/(2)sqrt(1-x^(2)):|_(0)^(1)+(1)/(2)sin^(-1)""(x)/(1):|_(0)^(1)]` `=0+sin^(-1)(1)=(pi)/(2)` sq. units. `"Total area "=((pi)/(2)+1)+(pi)/(2)=pi+1.` |
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