Question

Find the area under the curve of the following functions from limit 0 to π.

Functions are sin (x) , cos (x).​

Answer 1

$\large\underline{\sf{Solution-i}}$

Given curve is

$\rm :\longmapsto\:f(x) = sinx$

So, the required area bounded by the curve between x = 0 and x =  π with respect to x - axis is

$\rm \:  =  \:2 \displaystyle\int_0^{\dfrac{\pi}{2} } \: sinx \: dx$

$\rm \:  =  \: 2 \bigg| - cosx\bigg|_0^{\dfrac{\pi}{2} }$

$\rm \:  =  \: 2\bigg | - cos\dfrac{\pi}{2} + cos0\bigg|$

$\rm \:  =  \: 2\bigg | - 0 + 1\bigg|$

$\rm \:  =  \: 2\bigg | 1\bigg|$

$\rm \:  =  \: 2$

$\large\underline{\sf{Solution-ii}}$

Given curve is

$\rm :\longmapsto\:f(x) = cosx$

So, the required area bounded by the curve between x = 0 and x =  π with respect to x - axis is

$\rm \:  =  \:2 \displaystyle\int_0^{\dfrac{\pi}{2} } \: cosx \: dx$

$\rm \:  =  \: 2 \bigg|sinx\bigg|_0^{\dfrac{\pi}{2} }$

$\rm \:  =  \: 2 \bigg|sin \dfrac{\pi}{2} - sin0 \bigg|$

$\rm \:  =  \: 2 \bigg|1 - 0 \bigg|$

$\rm \:  =  \: 2 \times \bigg|1 \bigg|$

$\rm \:  =  \: 2$

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More to know :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$