Area Question 36
Question 36
- If
is a continuous function given by
Then, find the area of the region in the third quadrant bounded by the curves
Show Answer
Solution:
- Given,
Next, we have to find the points
$$ \begin{aligned} & =\int_{-2}^{-1 / 8} \sqrt{\frac{-x}{2}} d x-\int_{-2}^{-1}\left(x^{2}+2 x-1\right) d x-\int_{-1}^{-1 / 8} 2 x d x \ & =-\frac{2}{3 \sqrt{2}}\left[(-x)^{3 / 2}\right]{-2}^{-1 / 8}-\frac{x^{3}}{3}+x^{2}-x-{-2}^{-\left[x^{2}\right]_{-1}^{-1 / 8}} \ & =-\frac{2}{3 \sqrt{2}} \frac{1^{3 / 2}}{8}-2^{3 / 2}–\frac{1}{3}+1+1 \ & +-\frac{8}{3}+4+2-\frac{1}{64}-1 \ & =\frac{\sqrt{2}}{3}\left[2 \sqrt{2}-2^{-9 / 2}\right]+\frac{5}{3}+\frac{63}{64} \ & =\frac{63}{16 \times 3}+\frac{509}{64 \times 3}=\frac{761}{192} \text { squnits } \end{aligned} $$