Integration

J

James2

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Oct 14, 2009
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Hi, I hope this makes sense there are two questions and what I think may be the answers. I would like to know if I'm on the right lines please.
Q1)

If v = 5 + 24t – 3t2 ( v = m/s )

Calculate using integration what is the distance traveled during the interval
t = 1s & t = 6s

A1)

dy = [sup:1i5bm4wd]6[/sup:1i5bm4wd]1( 5 + 12 * x[sup:1i5bm4wd]2[/sup:1i5bm4wd] – x[sup:1i5bm4wd]2[/sup:1i5bm4wd] + c )
dx
= 385 m/s

Q2)

If y = cos2? where ? varies between ? =0 and ? = ?/4 radians

Using integration what is the area of a template.

A2)

dy = [sup:1i5bm4wd]1[/sup:1i5bm4wd]0( ¼ sin2? + c )
dx
= 0.71416

Thanks James
 
For question one, your answer is in m/s when the problem is asking for distance (in m. alone).

What is nice about velocity and distance traveled is this: v(t) = dx/dt

Notice if you integrate both sides, the right side becomes x(t), the function that relates distance traveled and time.
 
If v = 5 + 24t – 3t2 ( v = m/s )

Calculate using integration what is the distance traveled during the interval
t = 1s & t = 6s

61( 5 + 12 * x2 – x2 + c ) = 385 m/s
Click to expand...

The integral of v(t) = 5 + 24t – 3t^2 is 5t + 12t^2 – t^3 + C. Evaluating with limits of integration of 1 to 6, the displacement (which is not always the same as distance but is in this case) is 230 m.
 
Thankyou for your replies, the m/s was a typing error. Now I can see the answer to Q1 it does make sense, but I'm still not sure about Q2.

Thanks again

James
 
\(\displaystyle s(t ) \ is \ the \ position \ function \ and \ v(t) \ = \ s'(t), \ hence \ s(t) \ = \ \int v(t)dt, \ (FTC \ II)\)


\(\displaystyle A1) \ s(t) \ = \ \int_{1}^{6}[5+24t-3t^2}]dt \ = \ 230m\)

\(\displaystyle A2) \ A \ = \ \int_{0}^{\pi/4}cos(2\theta)d\theta \ = \ \frac{1}{2} \ sq. \ units\)
 
\(\displaystyle If \ you \ don't \ know \ how \ to \ integrate \ elementary \ functions, \ then \ you \ are \ ahead \ of \ yourself.\)

\(\displaystyle Back \ track \ to \ a \ more \ primitive \ existence.\)
 
\(\displaystyle A \ = \ \int_{0}^{\pi/4}cos(2\theta)d\theta \ = \ \frac{sin(2\theta)}{2}\bigg]_{0}^{\pi/4} \ = \ sin(\theta)cos(\theta)\bigg]_{0}^{\pi/4}\)

\(\displaystyle = \ \bigg(\frac{1}{\sqrt2}\bigg)\bigg(\frac{1}{\sqrt2}\bigg)-(0)(1) \ = \ \frac{1}{2}.\)
 
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