Question

$\int\limits^\pi _0 {sinx} \, dx$

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Answer 1

Topic :- Basic Maths

$\maltese\:\underline{\textsf{\textbf{AnsWer :}}}\:\maltese$

Definite Integral

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$\longrightarrow\:\:\sf\displaystyle \sf \int\limits_{0}^{\pi} \sin(x) \, dx$

$\longrightarrow\:\:\sf\displaystyle \sf \Bigg[ - \cos(x) \Bigg]^{\pi}_{0}$

$\longrightarrow\:\:\sf\displaystyle \sf \Bigg[ - \cos(\pi) - \bigg( - \cos(0) \bigg) \Bigg]$

$\longrightarrow\:\:\sf\displaystyle \sf \Bigg[ - \cos(\pi) + \cos(0) \Bigg]$

$\longrightarrow\:\:\sf\displaystyle \sf \Bigg[ - \cos(180^{\circ}) + \cos(0) \Bigg]$

$\longrightarrow\:\:\sf\displaystyle \sf- ( - 1) + 1$

$\longrightarrow\:\:\sf\displaystyle \sf 1 + 1$

$\longrightarrow\: \: \underline{ \boxed{ \sf\displaystyle \sf 2}}$

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$\footnotesize \bullet\displaystyle \sf \int\limits { x}^{n} \, dx =\frac{ {x}^{n + 1} }{n + 1} + c \\\\\footnotesize\bullet\displaystyle \sf \int\limits {e}^{x} \, dx = {e}^{x} + c \\\\\footnotesize\bullet\displaystyle \sf \int\limits \frac{1}{ x } \, dx = ln(x) + c \\ \\ \footnotesize\bullet\displaystyle \sf \int\limits \sin(x) \, dx = - \cos(x) + c \\ \\ \footnotesize\bullet\displaystyle \sf \int\limits \cos(x) \, dx = \sin(x) + c \\ \\ \footnotesize\bullet\displaystyle \sf \int\limits \sec^{2} (x) \, dx = \tan(x) + c$

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$\boxed{\boxed{\begin{minipage}{4cm}\displaystyle\circ\sf\:\int{1\:dx}=x+c\\\\\circ\sf\:\int{a\:dx}=ax+c\\\\\circ\sf\:\int{x^n\:dx}=\dfrac{x^{n+1}}{n+1}+c\\\\\circ\sf\:\int{sin\:x\:dx}=-cos\:x+c\\\\\circ\sf\:\int{cos\:x\:dx}=sin\:x+c\\\\\circ\sf\:\int{sec^2x\:dx}=tan\:x+c\\\\\circ\sf\:\int{e^x\:dx}=e^x+c\end{minipage}}}$

Answer 2

Given :   $\int\limits^\pi_0 {\sin x} \, dx$

To Find :  Solve

Solution:

let say

 $y=\int\limits^\pi_0 {\sin x} \, dx$

on integration

 $y=\left[ {-\cos x} \right]^\pi_0$

=> y  = - ( cosπ  - cos0 )

=> y =  - (-1 -  1)

=> y = 2

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