Question

Using the definition of integral as the limit of

a sum, show that

a

0

sin x dx 1 cos a. = − ∫​

Answer 1

Appropriate Question :-

Using the definition of integral as the limit of a sum, show that $\displaystyle\int_{0}^{a}\rm\: sinx dx = 1 - cosa$

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

Given integral is

$\rm \: \displaystyle\int_{0}^{a}\rm\: sinx dx$

On comparing with $\displaystyle\int_{a}^{b}\rm\: f(x) dx$, we get

$\rm \: a = 0$

$\rm \: b = a$

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

So,

$\rm \: nh \: = \: b - a - - - (1)$

Now, Using Definition of Limit as a Sum, we have

$\rm \: \displaystyle\int_{a}^{b}f(x) \: dx = \displaystyle\lim_{h \to 0}h\sum_{r=0}^{n-1}\rm f(a + rh)$

$\rm \: \displaystyle\int_{0}^{a}\rm \: sinx \: dx = \displaystyle\lim_{h \to 0}h\sum_{r=0}^{n-1}\rm f(rh)$

So, on substituting the values, we get

$\rm \: \displaystyle\int_{0}^{a} \rm \: sinx \: dx = \displaystyle\lim_{h \to 0}h\sum_{r=0}^{n-1}\rm sin(rh)$

$\rm \: = \displaystyle\lim_{h \to 0}\rm h[sin0 + sinh + sin2h+ \cdots + sin(n - 1)h]$

We know,

$\rm sina + sin(a + b) + sin(a + 2b) + \cdots + sin(a + (n - 1)b) \\ \rm \: = \: \dfrac{sin\bigg(\dfrac{nb}{2} \bigg)sin\bigg(a + \dfrac{(n - 1)b}{2} \bigg) }{sin\dfrac{b}{2} }$

So, using this result, we get (here a = 0 and b = h)

$\rm \: = \displaystyle\lim_{h \to 0}\rm h \times \dfrac{sin\bigg(\dfrac{nh}{2} \bigg)sin\bigg(0 + \dfrac{(n - 1)h}{2} \bigg) }{sin\dfrac{h}{2} }$

can be rewritten as

$\rm \: = \displaystyle\lim_{h \to 0}\rm \: sin\bigg(\dfrac{nh}{2} \bigg)sin\bigg(\dfrac{nh - h}{2} \bigg) \times \displaystyle\lim_{h \to 0}\rm \: \frac{h}{sin \dfrac{h}{2} }$

$\rm \: = \displaystyle\lim_{h \to 0}\rm \: sin\bigg(\dfrac{a}{2} \bigg)sin\bigg(\dfrac{a - h}{2} \bigg) \times 2\displaystyle\lim_{h \to 0}\rm \: \frac{\dfrac{h}{2} }{sin \dfrac{h}{2} }$

$\rm \: = \: sin\bigg(\dfrac{a}{2} \bigg)sin\bigg(\dfrac{a}{2} \bigg) \times 2 \times 1$

$\rm \: = \: 2sin^{2} \bigg(\dfrac{a}{2} \bigg)$

$\rm \: = \: 1 - cosa$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\rm \: \displaystyle\int_{0}^{a}\rm\: sinx dx \: = \: 1 \: - \: cosa \: }}$

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Additional Information :-

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

Answer 2

$\small\colorbox{lightyellow} {\text{ \bf♕ Brainliest answer }}$

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$\mathbb\red{ \tiny A \scriptsize \: N \small \:S \large \: W \Large \:E \huge \: R}$

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Here a=0, b=a, a=nh, f(x)=sin x .

$\[ \begin{aligned} \therefore \quad &\rm \int_{0}^{a} \sin x d x=\lim _{n \rightarrow x} \sum_{r=1}^{\infty}(a / n) \sin (r a / n) \\ \\ &\rm=\lim _{n \rightarrow \infty} \frac{a}{n}\left[\sin \frac{a}{n}+\sin \frac{2 a}{n}+\ldots+\frac{\sin n a}{n}\right] \\\\ &\rm=\lim _{n \rightarrow \infty} \frac{a}{n} \frac{\sin \left(\frac{a}{n}+\frac{n-1}{2} \cdot \frac{a}{n}\right) \sin \frac{n a}{2 n}}{\sin \frac{a}{2 n}} \\ \\ &\rm=\lim _{n \rightarrow \infty} \frac{2(a / 2 n)}{\sin (a / 2 n)} \cdot \sin \left(\frac{1}{2} a\left(1+\frac{1}{n}\right)\right) [\sin \left(\frac{a}{2}\right) \\\\ &\rm=2\cdot 1 \sin \left(\frac{a}{2}\right) \sin \left(\frac{a}{2}\right)\\\\ &\rm=2 \sin ^{2}\left(\frac{a}{2}\right)\\\\ &\rm=1-\cos a \end{aligned} \]$