Recent content by Jude075

So if I was to find the indefinite integral, both answers were right?

Homework Statement ∫ Sinx cos x dx Homework Equations The Attempt at a Solution If you integrate it using substitution, you get -cos2(x)/2but if you use double angle formula to rewrite the problem, it will be ∫1/2 sin(2x), and integrate it, you get -cos(2x)/4. isn't it weird?

So how do you guys use that expression to find its area? Do I derive it or integrate it?

Homework Statement A rectangle has its base on the x-axis and its vertices on the positive portion of the parabola $$ y=2-3x^2 $$ What is the maximum possible area of this rectangle? A. (8/27)*181/2 B.(2/9)*181/2 C. (4/15)*301/2...

You need to plug in f-1(x) instead of x, which is 0 .

Good explanation! Thank you very much!

I didn't know you could just factor 2 out of the sin. Is there any properties or laws for this?

$$\lim_{x \to 0} \frac{4\sin^2(2x)}{x^2}$$ How do you use that limit to get 16? This question comes from the AP test. So if I were to use the way I did to solve it , it would waste lots of time.

It's 0,so the slope is -6 :) Thank you guys very much!

This is just as difficult as finding the full expression for me. Please elaborate it more. Thank you!

Right! That is what made me wrong. Thank you so much! But I got the answer without using double-angle formula. Just keep applying L'hopital's rule:)

Homework Statement The function f(x)=5x2+3e2x is invertible. Give the slope of the normal line to the graph of f-1 at x=3 Homework Equations (f-1)'(x)=1/f'(f-1(x)) The slope of the normal line × the slope of the tangent =-1 So basically I just need to find -f'(f-1(x)) at x=3 The Attempt at a...

Okay. Here is how I tried. Lim (4*(sin x)2)/x2 because no much difference between 2/t and 1/t X→0 Then I got an indeterminant form 0/0, so I used L'hopital's rule. Lim (8sin x cos x)/(2x) X→0 Still 0/0 ,so again Lim (8(cosx)2-8(sinx)2)/2 X→0 Which equals 4. But my calculator says...

I still can't do it :(

Why did you change t→∞ to x→0? I don't understand this.