Area Question 36

Question 36

  1. If f(x) is a continuous function given by

f(x)=2x,|x|1 x2+ax+b,|x|>1

Then, find the area of the region in the third quadrant bounded by the curves x=2y2 and y=f(x) lying on the left on the line 8x+1=0.

(1999,5 M)

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Solution:

  1. Given, f(x)=2x, x2+ax+b,|x|>1

f(x)=x2+ax+b, if x<1 2x, if 1x<1 x2+ax+b, if x1

f is continuous on R, so f is continuous at -1 and 1 .

limx1f(x)=limx1+f(x)=f(1)  and limx1f(x)=limx1+f(x)=f(1) 1a+b=2 and 2=1+a+b ab=3 and a+b=1 a=2,b=1  Hence, f(x)=x2+2x1, if x<1 2x, if 1x<1 x2+2x1, if x1

Next, we have to find the points x=2y2 and y=f(x). The point of intersection is (2,1).

Required area =21/8x2f(x)dx

$$ \begin{aligned} & =\int_{-2}^{-1 / 8} \sqrt{\frac{-x}{2}} d x-\int_{-2}^{-1}\left(x^{2}+2 x-1\right) d x-\int_{-1}^{-1 / 8} 2 x d x \ & =-\frac{2}{3 \sqrt{2}}\left[(-x)^{3 / 2}\right]{-2}^{-1 / 8}-\frac{x^{3}}{3}+x^{2}-x-{-2}^{-\left[x^{2}\right]_{-1}^{-1 / 8}} \ & =-\frac{2}{3 \sqrt{2}} \frac{1^{3 / 2}}{8}-2^{3 / 2}–\frac{1}{3}+1+1 \ & +-\frac{8}{3}+4+2-\frac{1}{64}-1 \ & =\frac{\sqrt{2}}{3}\left[2 \sqrt{2}-2^{-9 / 2}\right]+\frac{5}{3}+\frac{63}{64} \ & =\frac{63}{16 \times 3}+\frac{509}{64 \times 3}=\frac{761}{192} \text { squnits } \end{aligned} $$



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