Graphs of two function f(x)=sin x and (g)x=cos x is given below: … - Vidyadip

Question

Graphs of two function $\text{f}(\text{x})=\text{sin}\text{ x}$ and $\text{(g)}\text{x}=\text{cos}\text{ x}$ is given below:

Based on the above information, answer the following questions.

In $(0, \pi)$, the curves $\text{f}(\text{x})=\text{sin}\text{ x}$ and $\text{g}\text{ (x)}=\text{cos}\text{ x}$ at $\text{x}=$
1. $\frac{\pi}{2}$
2. $\frac{\pi}{3}$
3. $\frac{\pi}{4}$
4. ${\pi}$
Value of $\int\limits_{0}^{\frac{\pi}{4}}\text{sin}\text{ x}\text{ dx}$ is.
1. $1-\frac{1}{\sqrt{2}}$
2. $1+\frac{1}{\sqrt{2}}$
3. $2-\frac{1}{\sqrt{2}}$
4. $2+\frac{1}{\sqrt{2}}$

Value of $\int\limits_\frac{\pi}{4}^{\frac{\pi}{2}}\text{cos}\text{ x}\text{ dx}$ is.
1. $1+\frac{1}{\sqrt{2}}$
2. $1-\frac{1}{\sqrt{2}}$
3. $2-\sqrt{2}$
4. $2+\sqrt{2}$
Value of $\int\limits_{0}^{\pi}\text{sin}\text{ x}\text{ dx}$ is.

0
1
2
-2

Value of $\int\limits_{0}^\frac{\pi}{2}\text{sin}\text{ x}\text{ dx}$ is.

0
1
3
4

✓

Answer

(c) $\frac{\pi}{4}$

Solution:

for point of intersection, we have

$\text{sin}\text{ x}=\text{cos}\text{ x}$

$\Rightarrow\frac{\text{sin}\text{ x}}{\text{cos}\text{ x}}=1$

$\Rightarrow\text{tan}\text{ x}=1$

$\Rightarrow\text{x}=\frac{\pi}{4}$

(a) $1-\frac{1}{\sqrt{2}}$

Solution:

$\int\limits_{0}^{\frac{\pi}{4}}\text{sin}\text{ x}\text{ dx}=\big[-\text{cos x}\big]^\frac{\pi}{4}_0=-\text{cos}\frac{\pi}{4}+\text{cos}0$

$=1-\frac{1}{\sqrt{2}}$

(b) $1-\frac{1}{\sqrt{2}}$

Solution:

$\int\limits_{\frac{\pi}{2}}^{\frac{\pi}{4}}\text{cos}\text{ x}\text{ dx}=\big[\text{sin x}\big]^\frac{\pi}{2}_\frac{\pi}{4}=\text{sin}\frac{\pi}{2}-\text{sin}\frac{\pi}{4}$

$=1-\frac{1}{\sqrt{2}}$

(c) 2

Solution:

$\int\limits_{0}^{\pi}\text{sin}\text{ x}\text{ dx}=\big[-\text{cos x}\big]^{\pi}_0=\big[-\text{cos}{\pi}+\text{cos}0\big]=2$

(b) 1

Solution:

$\int\limits_{0}^\frac{\pi}{2}\text{sin}\text{ x}\text{ dx}=\big[-\text{cos x}\big]^\frac{\pi}{2}_0=\Big[-\text{cos}\frac{\pi}{2}+\text{cos}0\Big]$

$=0+1+1$

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10