Find f. Antiderivatives (Our class hasn't used integrals yet).

Question

zenmo · Answer

$$f(t)=-4sint+tant$$
$$f(\pi/3)=-4\sin(\pi/3)+\tan(\pi/3)$$
$$=-4 \times \frac{ \sqrt{3} }{ 2 }+\sqrt{3}+C=3$$
$$=-2\sqrt{3}+\sqrt{3}+C=3$$
$$C=3+\sqrt{3}$$
Answer: $$f(t)= -4\sin(t) + \tan(t) + 3 + \sqrt{3}$$

zenmo · Answer

#9, correct picture.

solomonzelman · Answer

$\large\color{#000000 }{ \displaystyle f'(t)=4\cos(t)+\sec^2(t) }$

and you want to find f(t), correct?

solomonzelman · Answer

4sin(t) , and not negative

d/dx sin(x) = cos(x)
And consequentialy, the integral/antiderivative of cos(t) is sin(t).

zenmo · Answer

$$f(t)=4\sin(t)+tant+3-3\sqrt{3}$$ be the answer?

solomonzelman · Answer

I am not sure about 3-3√3 part, but the first two terms are right

solomonzelman · Answer

$\large\color{#000000 }{ \displaystyle f'(t)=4\sin(t)+\tan (t) +C}$

is the genera antiderivative.

zenmo · Answer

in the picture, there are information provided to find C as well.

solomonzelman · Answer

You know that  f(π/3)=3.        Use this to solve for C.

$\large\color{#000000 }{ \displaystyle \color{red}{3}=4\sin(\color{red}{\pi/3})+\tan (\color{red}{\pi/3}) +C}$

zenmo · Answer

$$f(\pi/3)=4\sin(\pi/3)+\tan(\pi/3)+C=3$$
$$4 \times \frac{ \sqrt{3} }{2 }+\sqrt{3}+C=3$$
$$2\sqrt{3}+1\sqrt{3}+C=3$$
$$3\sqrt{3}+C=3$$
$$C=3=3$$

zenmo · Answer

$$C=3-3\sqrt{3}$$**

solomonzelman · Answer

$\large\color{#000000 }{ \displaystyle 2\sqrt{3}+\sqrt{3}+C=3 }$
$\large\color{#000000 }{ \displaystyle 3\sqrt{3}+C=3 }$
$\large\color{#000000 }{ \displaystyle C=3 -3\sqrt{3} }$

yes your C is correct

zenmo · Answer

yep, that is the answer. Thanks!

solomonzelman · Answer

Anytime...