Master the Net Change Theorem with Expert Homework Statement Help

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Homework Statement


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Homework Equations





The Attempt at a Solution


i got -234 for the last part but it should be abs.value so 234.. but it didnt work.
is my work correct?
 
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No. You have v(t) = (t - 8)(t + 3), which is correct, but it's not true that v(t) <= 0 on the interval [6, 11]. Where did the 11 come from? Think about what the factors of v(t) mean. v(t) = 0 only for t = -3 and t = 8. These two numbers divide the number line into three regions, on two of which v(t) > 0 and on one of which v(t) < 0.
 
v(t) > 0 on [-infinity,-3] and [8, infinity]
v(t) < 0 on [-3, 8]

so how do i express the antiderivative?
 
\int_6^{12} |v(t)|dt~=~\int_6^8 -v(t) dt~+~\int_8^{12} v(t) dt

Isn't that what you're trying to do?
 
that what i didnt understand how to set up..
so pretty much you take the number you got for t (which was t= -3, 8)
you pick the one that in the interval [6, 12] (in this case, 8)
and then make it 6 -- > 8 + 8 --> 12
i got it. sweet tnx a lot!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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