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How would I answer this? 

 Apr 28, 2020

Best Answer 

 #2
avatar+24983 
+2

Integration

Suppose that \(f'(x) = \dfrac{4}{\sqrt{1-x^2}}\) and that \(f(0) = 2\).
Find \(f\left(\dfrac{\sqrt{3}}{2}\right)\).

 

\(\begin{array}{|rcll|} \hline && \mathbf{\large{ \int } \dfrac{4}{\sqrt{1-x^2}} \ dx} \\\\ &=& 4\mathbf{\large{ \int } \dfrac{1}{\sqrt{1-x^2}} \ dx} \qquad \boxed{ \text{substitute: } \\ x = \sin(\theta),\ dx=\cos(\theta)d\theta } \\\\ &=& 4 \large{ \int } \dfrac{\cos(\theta)}{ \sqrt{1-\Big(\sin(\theta)\Big)^2} }\ d\theta \\\\ &=& 4 \displaystyle \int \dfrac{\cos(\theta)}{ \cos(\theta) }\ d\theta \\\\ &=& 4 \displaystyle \int d\theta \\ &=& 4 \theta + c \quad | \quad \theta = \arcsin(x) \\ &=& 4 \Big(\arcsin(x)\Big) + c \\ \hline f(x) &=& 4 \Big(\arcsin(x)\Big) + c \quad | \quad x=0 \\ f(0) &=& 4 \Big(\arcsin(0)\Big) + c \quad | \quad f(0) = 2 \\ 2 &=& 4 \Big(\arcsin(0)\Big) + c \\ 2 &=& 4 \cdot 0 + c \\ 2 &=& 0 + c \\ \mathbf{c} &=& \mathbf{2} \\ \hline \mathbf{f(x)} &=& \mathbf{4 \Big(\arcsin(x)\Big) + 2} \\\\ f\left(\dfrac{\sqrt{3}}{2}\right) &=& 4 \Big(\arcsin\left(\dfrac{\sqrt{3}}{2}\right)\Big) + 2 \\\\ f\left(\dfrac{\sqrt{3}}{2}\right) &=& 4\cdot \Big(\dfrac{\pi}{3}\Big) + 2 \\\\ f\left(\dfrac{\sqrt{3}}{2}\right) &=& \dfrac{4\pi}{3} + 2 \\\\ \mathbf{f\left(\dfrac{\sqrt{3}}{2}\right)} &=& \mathbf{6.18879020479} \\ \hline \end{array}\)

 

laugh

 Apr 29, 2020
 #1
avatar+23646 
+1

I think integral   1/(sqrt(1-x^2)     is a standard     arcsin (x)

 

4 arcsin x   + c        f(0) = 2   means   c = 2

 

4 arcsin (sqrt3 /2) + 2 = 242  o

 Apr 28, 2020
 #2
avatar+24983 
+2
Best Answer

Integration

Suppose that \(f'(x) = \dfrac{4}{\sqrt{1-x^2}}\) and that \(f(0) = 2\).
Find \(f\left(\dfrac{\sqrt{3}}{2}\right)\).

 

\(\begin{array}{|rcll|} \hline && \mathbf{\large{ \int } \dfrac{4}{\sqrt{1-x^2}} \ dx} \\\\ &=& 4\mathbf{\large{ \int } \dfrac{1}{\sqrt{1-x^2}} \ dx} \qquad \boxed{ \text{substitute: } \\ x = \sin(\theta),\ dx=\cos(\theta)d\theta } \\\\ &=& 4 \large{ \int } \dfrac{\cos(\theta)}{ \sqrt{1-\Big(\sin(\theta)\Big)^2} }\ d\theta \\\\ &=& 4 \displaystyle \int \dfrac{\cos(\theta)}{ \cos(\theta) }\ d\theta \\\\ &=& 4 \displaystyle \int d\theta \\ &=& 4 \theta + c \quad | \quad \theta = \arcsin(x) \\ &=& 4 \Big(\arcsin(x)\Big) + c \\ \hline f(x) &=& 4 \Big(\arcsin(x)\Big) + c \quad | \quad x=0 \\ f(0) &=& 4 \Big(\arcsin(0)\Big) + c \quad | \quad f(0) = 2 \\ 2 &=& 4 \Big(\arcsin(0)\Big) + c \\ 2 &=& 4 \cdot 0 + c \\ 2 &=& 0 + c \\ \mathbf{c} &=& \mathbf{2} \\ \hline \mathbf{f(x)} &=& \mathbf{4 \Big(\arcsin(x)\Big) + 2} \\\\ f\left(\dfrac{\sqrt{3}}{2}\right) &=& 4 \Big(\arcsin\left(\dfrac{\sqrt{3}}{2}\right)\Big) + 2 \\\\ f\left(\dfrac{\sqrt{3}}{2}\right) &=& 4\cdot \Big(\dfrac{\pi}{3}\Big) + 2 \\\\ f\left(\dfrac{\sqrt{3}}{2}\right) &=& \dfrac{4\pi}{3} + 2 \\\\ \mathbf{f\left(\dfrac{\sqrt{3}}{2}\right)} &=& \mathbf{6.18879020479} \\ \hline \end{array}\)

 

laugh

heureka Apr 29, 2020

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