HELP???????????????????????????????????????????????????????? pls? (10 points) (image down below to make it easier to understand.
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2 answers:
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The indefinite integral will be
When we integrate any function without the limits then it will be an indefinite integral.
General Formulas and Concepts:
Integration Rule [Reverse Power Rule]:
Integration Property [Multiplied Constant]:
Integration Property [Addition/Subtraction]:
[Integral] Rewrite [Integration Property - Addition/Subtraction]:
[Integrals] Rewrite [Integration Property - Multiplied Constant]:
[Integrals] Reverse Power Rule:
Simplify:
So the indefinite integral will be
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A)
why is 0? well, the bounds are the same.
b)
let's use the second fundamental theorem of calculus, where F'(x) = dF/du * du/dx
![\bf \displaystyle F(x)=\int\limits_{4}^{u}\cfrac{2t-1}{t+2}\cdot dt\qquad F'(x)=\cfrac{dF}{du}\cdot \cfrac{du}{dx} \\\\\\ \displaystyle\cfrac{d}{du}\left[ \int\limits_{4}^{u}\cfrac{2t-1}{t+2}\cdot dt \right]\cdot \cfrac{du}{dx}\implies \cfrac{2u-1}{u+2}\cdot \cfrac{1}{2\sqrt{x}} \\\\\\ \cfrac{2\sqrt{x}}{\sqrt{x}+2}\cdot \cfrac{1}{2\sqrt{x}}\implies \cfrac{2\sqrt{x}-1}{2x+4\sqrt{x}}](https://tex.z-dn.net/?f=%5Cbf%20%5Cdisplaystyle%20F%28x%29%3D%5Cint%5Climits_%7B4%7D%5E%7Bu%7D%5Ccfrac%7B2t-1%7D%7Bt%2B2%7D%5Ccdot%20dt%5Cqquad%20F%27%28x%29%3D%5Ccfrac%7BdF%7D%7Bdu%7D%5Ccdot%20%5Ccfrac%7Bdu%7D%7Bdx%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Cdisplaystyle%5Ccfrac%7Bd%7D%7Bdu%7D%5Cleft%5B%20%5Cint%5Climits_%7B4%7D%5E%7Bu%7D%5Ccfrac%7B2t-1%7D%7Bt%2B2%7D%5Ccdot%20dt%20%5Cright%5D%5Ccdot%20%5Ccfrac%7Bdu%7D%7Bdx%7D%5Cimplies%20%5Ccfrac%7B2u-1%7D%7Bu%2B2%7D%5Ccdot%20%5Ccfrac%7B1%7D%7B2%5Csqrt%7Bx%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7B2%5Csqrt%7Bx%7D%7D%7B%5Csqrt%7Bx%7D%2B2%7D%5Ccdot%20%5Ccfrac%7B1%7D%7B2%5Csqrt%7Bx%7D%7D%5Cimplies%20%5Ccfrac%7B2%5Csqrt%7Bx%7D-1%7D%7B2x%2B4%5Csqrt%7Bx%7D%7D)
c)
we know x = 16, we also know from section a) that at that point f(x) = y = 0, so the point is at (16, 0), using section b) let's get the slope,
d)
now, we can get critical points from zeroing out the derivative, we also get critical points from zeroing out the denominator, however, the ones from the denominator are points where the function is not differentiable, namely, is not a smooth curve, is a sharp jump, a cusp, or a spike, and therefore those points are usually asymptotic, however, they're valid critical points, let's check both,
now, doing a first-derivative test on those regions, we get the values as in the picture below.
so, you can see where is increasing and decreasing.
why is 0? well, the bounds are the same.
b)
let's use the second fundamental theorem of calculus, where F'(x) = dF/du * du/dx
c)
we know x = 16, we also know from section a) that at that point f(x) = y = 0, so the point is at (16, 0), using section b) let's get the slope,
d)
now, we can get critical points from zeroing out the derivative, we also get critical points from zeroing out the denominator, however, the ones from the denominator are points where the function is not differentiable, namely, is not a smooth curve, is a sharp jump, a cusp, or a spike, and therefore those points are usually asymptotic, however, they're valid critical points, let's check both,
now, doing a first-derivative test on those regions, we get the values as in the picture below.
so, you can see where is increasing and decreasing.