Help with the following questions please (1 Viewer)

[Boonyak]

Hey guys can you help with these questions mainly 8 and 9 if you could do 7 and 10 it would be appreciated, for question 8 i dont get how to sketch it showing pi/6 etc, please show with working out if you dont want to do the whole question please do the begging to help me start thanks


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[Boonyak]

Note ive done 6 thanks

 

[Timske]

<a href="http://www.codecogs.com/eqnedit.php?latex=\dpi{150} \int_{0}^{\frac{\pi}{3}}\textup{ sin}(2-\textup{x}) ~\textup{dx} \\ = \left [\textup{cos}(2-\textup{x}) \right ]_{0}^{\frac{\pi}{3}} \\ = cos(2-\frac{\pi}{3}) - cos(2) \\ \approx 0.9955 ~(4 dp)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\dpi{150} \int_{0}^{\frac{\pi}{3}}\textup{ sin}(2-\textup{x}) ~\textup{dx} \\ = \left [\textup{cos}(2-\textup{x}) \right ]_{0}^{\frac{\pi}{3}} \\ = cos(2-\frac{\pi}{3}) - cos(2) \\ \approx 0.9955 ~(4 dp)" title="\dpi{150} \int_{0}^{\frac{\pi}{3}}\textup{ sin}(2-\textup{x}) ~\textup{dx} \\ = \left [\textup{cos}(2-\textup{x}) \right ]_{0}^{\frac{\pi}{3}} \\ = cos(2-\frac{\pi}{3}) - cos(2) \\ \approx 0.9955 ~(4 dp)" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex=\dpi{100} \int_{0.5}^{1} \textup{sec}^2(\textup{x}) ~\textup{dx} , ~ integral ~of~ sec^2(x) ~is ~tan(x) \\\\ = \left [ \textup{tan}(\textup{x}) \right ]_{0.5}^{1} \\ = \textup{tan}(1) - \textup{tan}(0.5) \\ \approx 1.01 (2\textup{dp})" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\dpi{100} \int_{0.5}^{1} \textup{sec}^2(\textup{x}) ~\textup{dx} , ~ integral ~of~ sec^2(x) ~is ~tan(x) \\\\ = \left [ \textup{tan}(\textup{x}) \right ]_{0.5}^{1} \\ = \textup{tan}(1) - \textup{tan}(0.5) \\ \approx 1.01 (2\textup{dp})" title="\dpi{100} \int_{0.5}^{1} \textup{sec}^2(\textup{x}) ~\textup{dx} , ~ integral ~of~ sec^2(x) ~is ~tan(x) \\\\ = \left [ \textup{tan}(\textup{x}) \right ]_{0.5}^{1} \\ = \textup{tan}(1) - \textup{tan}(0.5) \\ \approx 1.01 (2\textup{dp})" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex=\dpi{120} A = \int_{a}^{b} y~dx \\ = 2 \int_{0}^{\frac{\pi}{6}} 3sinx ~dx \\ = 2\left [ -3cos(x) \right ]_{0}^{\frac{\pi}{6}} \\ =2\left [ -3cos(\frac{\pi}{6}) @plus; 3 \right ] \\ =2\left [ \frac{-3\sqrt{3}}{2} @plus; 3\right ] \\ =-3\sqrt{3} @plus; 6~ units^2" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\dpi{120} A = \int_{a}^{b} y~dx \\ = 2 \int_{0}^{\frac{\pi}{6}} 3sinx ~dx \\ = 2\left [ -3cos(x) \right ]_{0}^{\frac{\pi}{6}} \\ =2\left [ -3cos(\frac{\pi}{6}) + 3 \right ] \\ =2\left [ \frac{-3\sqrt{3}}{2} + 3\right ] \\ =-3\sqrt{3} + 6~ units^2" title="\dpi{120} A = \int_{a}^{b} y~dx \\ = 2 \int_{0}^{\frac{\pi}{6}} 3sinx ~dx \\ = 2\left [ -3cos(x) \right ]_{0}^{\frac{\pi}{6}} \\ =2\left [ -3cos(\frac{\pi}{6}) + 3 \right ] \\ =2\left [ \frac{-3\sqrt{3}}{2} + 3\right ] \\ =-3\sqrt{3} + 6~ units^2" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex=\dpi{120} 10 ~b) \int_{0}^{\frac{\pi}{2}} cos^3(x)sin(x)~dx = \left [ -\frac{1}{4}cos^4(x) \right ]_{0}^{\frac{\pi}{2}}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\dpi{120} 10 ~b) \int_{0}^{\frac{\pi}{2}} cos^3(x)sin(x)~dx = \left [ -\frac{1}{4}cos^4(x) \right ]_{0}^{\frac{\pi}{2}}" title="\dpi{120} 10 ~b) \int_{0}^{\frac{\pi}{2}} cos^3(x)sin(x)~dx = \left [ -\frac{1}{4}cos^4(x) \right ]_{0}^{\frac{\pi}{2}}" /></a>

 

[qwerty44]

 

[Boonyak]

Thanks for the quick and detailed responses highLy
Apreciated and just a question if I were
To sketch the area question where would pi/6 lie on the x axis ?

 

[Timske]

Thanks for the quick and detailed responses highLy
Apreciated and just a question if I were
To sketch the area question where would pi/6 lie on the x axis ?

well pi/6 is 30. so roughly between 0 and pi/2 more to the left. if that explains it...

 

[mkavadas]

can someone please help me with this question!!!!

 

[Pfortune35]

can someone please help me with this question!!!!
View attachment 25683

start a new thread, in the ecos or buissness section.

 

[twinklegal19]

<a href="http://www.codecogs.com/eqnedit.php?latex=\dpi{150} \int_{0}^{\frac{\pi}{3}}\textup{ sin}(2-\textup{x}) ~\textup{dx} \\ = \left [\textup{cos}(2-\textup{x}) \right ]_{0}^{\frac{\pi}{3}} \\ = cos(2-\frac{\pi}{3}) - cos(2) \\ \approx 0.9955 ~(4 dp)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\dpi{150} \int_{0}^{\frac{\pi}{3}}\textup{ sin}(2-\textup{x}) ~\textup{dx} \\ = \left [\textup{cos}(2-\textup{x}) \right ]_{0}^{\frac{\pi}{3}} \\ = cos(2-\frac{\pi}{3}) - cos(2) \\ \approx 0.9955 ~(4 dp)" title="\dpi{150} \int_{0}^{\frac{\pi}{3}}\textup{ sin}(2-\textup{x}) ~\textup{dx} \\ = \left [\textup{cos}(2-\textup{x}) \right ]_{0}^{\frac{\pi}{3}} \\ = cos(2-\frac{\pi}{3}) - cos(2) \\ \approx 0.9955 ~(4 dp)" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex=\dpi{100} \int_{0.5}^{1} \textup{sec}^2(\textup{x}) ~\textup{dx} , ~ integral ~of~ sec^2(x) ~is ~tan(x) \\\\ = \left [ \textup{tan}(\textup{x}) \right ]_{0.5}^{1} \\ = \textup{tan}(1) - \textup{tan}(0.5) \\ \approx 1.01 (2\textup{dp})" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\dpi{100} \int_{0.5}^{1} \textup{sec}^2(\textup{x}) ~\textup{dx} , ~ integral ~of~ sec^2(x) ~is ~tan(x) \\\\ = \left [ \textup{tan}(\textup{x}) \right ]_{0.5}^{1} \\ = \textup{tan}(1) - \textup{tan}(0.5) \\ \approx 1.01 (2\textup{dp})" title="\dpi{100} \int_{0.5}^{1} \textup{sec}^2(\textup{x}) ~\textup{dx} , ~ integral ~of~ sec^2(x) ~is ~tan(x) \\\\ = \left [ \textup{tan}(\textup{x}) \right ]_{0.5}^{1} \\ = \textup{tan}(1) - \textup{tan}(0.5) \\ \approx 1.01 (2\textup{dp})" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex=\dpi{120} A = \int_{a}^{b} y~dx \\ = 2 \int_{0}^{\frac{\pi}{6}} 3sinx ~dx \\ = 2\left [ -3cos(x) \right ]_{0}^{\frac{\pi}{6}} \\ =2\left [ -3cos(\frac{\pi}{6}) @plus; 3 \right ] \\ =2\left [ \frac{-3\sqrt{3}}{2} @plus; 3\right ] \\ =-3\sqrt{3} @plus; 6~ units^2" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\dpi{120} A = \int_{a}^{b} y~dx \\ = 2 \int_{0}^{\frac{\pi}{6}} 3sinx ~dx \\ = 2\left [ -3cos(x) \right ]_{0}^{\frac{\pi}{6}} \\ =2\left [ -3cos(\frac{\pi}{6}) + 3 \right ] \\ =2\left [ \frac{-3\sqrt{3}}{2} + 3\right ] \\ =-3\sqrt{3} + 6~ units^2" title="\dpi{120} A = \int_{a}^{b} y~dx \\ = 2 \int_{0}^{\frac{\pi}{6}} 3sinx ~dx \\ = 2\left [ -3cos(x) \right ]_{0}^{\frac{\pi}{6}} \\ =2\left [ -3cos(\frac{\pi}{6}) + 3 \right ] \\ =2\left [ \frac{-3\sqrt{3}}{2} + 3\right ] \\ =-3\sqrt{3} + 6~ units^2" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex=\dpi{120} 10 ~b) \int_{0}^{\frac{\pi}{2}} cos^3(x)sin(x)~dx = \left [ -\frac{1}{4}cos^4(x) \right ]_{0}^{\frac{\pi}{2}}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\dpi{120} 10 ~b) \int_{0}^{\frac{\pi}{2}} cos^3(x)sin(x)~dx = \left [ -\frac{1}{4}cos^4(x) \right ]_{0}^{\frac{\pi}{2}}" title="\dpi{120} 10 ~b) \int_{0}^{\frac{\pi}{2}} cos^3(x)sin(x)~dx = \left [ -\frac{1}{4}cos^4(x) \right ]_{0}^{\frac{\pi}{2}}" /></a>

View attachment 25680

When you integrate sinx, don't you get negative cosx?

 

[SpiralFlex]

^Division by -1.

 

[twinklegal19]

^Division by -1.

oh woops sorry, I take that back Timske

didn't realise the negative coefficient of x

/fail