Question

please answer the question fast ​

Answer 1

EXPLANATION.

$\implies \displaystyle \int\limits^\pi_0 {cos(x)} \, dx$

As we know that,

⇒ ∫cos x dx = sin x + c.

Using this formula in the equation, we get.

$\implies \displaystyle \bigg[ sin x \bigg]_{0}^{\pi}$

As we know that,

In definite integration first we put the upper limit then we put the lower limit, we get.

$\implies \displaystyle \bigg[sin(\pi) - sin(0) \bigg]$

As we know that,

⇒ sin(π) = 0.

⇒ sin(0) = 0.

$\implies \displaystyle \bigg[sin(\pi) - sin(0) \bigg] = 0.$

                                                                                                                         

MORE INFORMATION.

Summation of series by integration.

For finding sum of an infinite series with the help of definite integration, following formula be used,

$\sf \implies \displaystyle \lim_{n \to \infty} \sum_{r = 0}^{n - 1} f \bigg( \dfrac{r}{n} \bigg) . \dfrac{1}{n} = \int\limits^1_0 {f(x)} \, dx$

Working rules.

(1) = Express the given series in the form of,

$\sf \implies \displaystyle \lim_{n \to \infty} \sum_{r = 0}^{n - 1} f \bigg( \dfrac{r}{n} \bigg) . \dfrac{1}{n}$

(2) = Replace (r/n) by x, 1/n by dx, and ∑ by ∫, we get the integral ∫f(x)dx in place of above series.

(3) = The lower limit of this integral = $\sf \displaystyle \lim_{n \to \infty} \bigg( \dfrac{r}{n} \bigg)_{r = 0}$  r is least value in this case r = 0.

(4) = Upper limit = $\sf \implies \displaystyle \lim_{n \to \infty} \bigg( \dfrac{r}{n} \bigg)_{r = n - 1}$  r is greatest values in this case r = n - 1.

Answer 2

Given

$\int_{0}^{\pi} cosxdx$

Remember

1) Remember the rule that you have learnt in integration.

$\int{sinxdx} = -cosx+C$

Similarly,

$\int{cosxdx} = sinx+C$

Since the limits of integration is given, we will not take $C$.

2) The value of radian $\pi$ is assumed to be 180°.

Solution

$\int_{0}^{\pi} cosxdx$

$=\left[sinx\right]^\pi_0$

By substituting the radian and 0 in place of x, we get:

$=sin \pi - sin0 \degree$

If you remember about the trigonometric table, you get to know that:

$\implies sin180 = 0$

$\implies sin0 = 0$

Therefore,

$\int_{0}^{\pi} cosxdx = 0$

Hope it helps you!