Let f be the continuous function defined on [−4, 3]
whose graph, consisting of three line segments and a
semicircle centered at the origin, is given above. Let g
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OpenStudy (ksaimouli):
\[g(x)=\int\limits_{1}^{x}f(t)dt\]
OpenStudy (ksaimouli):
find g(-2)
OpenStudy (ksaimouli):
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OpenStudy (ksaimouli):
@hartnn
OpenStudy (anonymous):
Solve this I have a container with 4,200 milliliters of water. How many litres are in my container?
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OpenStudy (ksaimouli):
i need to find g(-2)
OpenStudy (anonymous):
the equation is the area of a triangle minus the area of a semi circle... do you see that?
OpenStudy (anonymous):
the traingle has height = 3 and base = 1, so A(triangle)=1/2 baseXheight = (1/2)(1)(3) = 3/2
OpenStudy (anonymous):
the semi circle has area of (1/2)PiR^2 where R=1, That means the area = Pi/2
OpenStudy (ksaimouli):
g(2)
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OpenStudy (anonymous):
\[\int\limits_{1}^{-2} = -\int\limits_{-2}^{1} \]
OpenStudy (ksaimouli):
yup
OpenStudy (ksaimouli):
i got it thx what about g(2)
OpenStudy (ksaimouli):
usind distance formula?
OpenStudy (anonymous):
Im thinking there might be an easier way,
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OpenStudy (ksaimouli):
ok
OpenStudy (anonymous):
\[(\frac{ -1-1 }{ 3-1 })(1)\] will give you the height of the triangle
OpenStudy (ksaimouli):
how did u get that
OpenStudy (anonymous):
then the base is 1
OpenStudy (ksaimouli):
@Edutopia how did u get that ?
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OpenStudy (ksaimouli):
@Edutopia
OpenStudy (anonymous):
yeah, it should be (-1-0)/(3-1), my mistake: that is the slope which is the change in y with respect to x, so as x change one (the distance between 1 and 2) y will change -1/2
OpenStudy (ksaimouli):
what points did u choose
OpenStudy (anonymous):
A(triangle)=(-1/2)(1)(1/2), i used the two given points you have above in your graph
OpenStudy (ksaimouli):
ok
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