The area bounded by y=x+1, y=cos x and the X- | vedclass

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(B) The curves are $y=x+1$ and $y=\cos x$. The region is bounded by these curves and the $X$-axis.From the graph, the region is split into two parts a

Vedclass · Accepted Answer

$\frac{3}{2} 	ext{ sq unit}$

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(B) The curves are $y=x+1$ and $y=\cos x$. The region is bounded by these curves and the $X$-axis.From the graph, the region is split into two parts at $x=0$.For $x \in [-1, 0]$, the region is bounded by $y=x+1$ and the $X$-axis.For $x \in [0, \pi/2]$, the region is bounded by $y=\cos x$ and the $X$-axis.$	ext{Required Area} = \int_{-1}^{0} (x+1) dx + \int_{0}^{\pi/2} \cos x dx$$= \left[ \frac{x^2}{2} + x ight]_{-1}^{0} + [\sin x]_{0}^{\pi/2}$$= \left( (0) - \left( \frac{(-1)^2}{2} - 1 ight) ight) + (\sin(\pi/2) - \sin(0))$$= \left( 0 - (\frac{1}{2} - 1) ight) + (1 - 0)$$= \frac{1}{2} + 1 = \frac{3}{2} 	ext{ sq units.}$

The area bounded by $y=x+1$, $y=\cos x$ and the $X$-axis is

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