Mathematics, 25.08.2019 01:30 pineapplepizaaaaa

Ex 3.7
12. find the area between the curve y=x³-2 and the y-axis between y= -1 and y=25

Answers: 2

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$y=x^3-2\\
x^3=y+2\\
x=\sqrt[3]{y+2}\\\\
\int \limits_{-1}^{25}\sqrt[3]{y+2}\, dy=\\
\int \limits_{-1}^{25}(y+2)^{\tfrac{1}{3}}\, dy=\\
\left[\dfrac{(y+2)^{\tfrac{4}{3}}}{\frac{4}{3}} \right]_{-1}^{25}=\\
$
$\left[\dfrac{3}{4}(y+2)^{\tfrac{4}{3}} \right]_{-1}^{25}=\\
\left[\dfrac{3}{4}(y+2)\sqrt[3]{y+2} \right]_{-1}^{25}=\\
\dfrac{3}{4}(25+2)\sqrt[3]{25+2}-\left(\dfrac{3}{4}(-1+2)\sqrt[3]{-1+2}\right)=\\
\dfrac{3}{4}\cdot27\sqrt[3]{27}-\left(\dfrac{3}{4}\sqrt[3]{1}\right)=\\
\dfrac{3}{4}\cdot27\cdot3-\dfrac{3}{4}=\\
\dfrac{3}{4}(81-1)=\\
\dfrac{3}{4}\cdot 80=\\3\cdot20=\\
60
$

Answer from: christingle2004

Yeah, you'd have to use the inverse function to produce this result.

Let's get the inverse function first:

$y={ x }^{ 3 }-2\\ \\ { x }^{ 3 }=y+2\\ \\ x=\sqrt [ 3 ]{ y+2 }$

$\\ \\ \therefore \quad { f }^{ -1 }\left( x \right) =\sqrt [ 3 ]{ x+2 }$

Now, we can solve the problem using:

$\int _{ -1 }^{ 25 }{ \sqrt [ 3 ]{ x+2 } } dx$

But to solve the problem more easily we make u=x+2, therefore du/dx=1, therefore du=dx.

When x=25, u=27.

When x=-1, u=1.

Now:

$\int _{ 1 }^{ 27 }{ { u }^{ \frac { 1 }{ 3 } } } du\\ \\ ={ \left[ \frac { 3 }{ 4 } { u }^{ \frac { 4 }{ 3 } } \right] }_{ 1 }^{ 27 }$

$\\ \\ =\frac { 3 }{ 4 } \cdot { 27 }^{ \frac { 4 }{ 3 } }-\frac { 3 }{ 4 } \cdot { 1 }^{ \frac { 4 }{ 3 } }\\ \\ =\frac { 3 }{ 4 } { \left( { 3 }^{ 3 } \right) }^{ \frac { 4 }{ 3 } }-\frac { 3 }{ 4 }$

$\\ \\ =\frac { 3 }{ 4 } \cdot { 3 }^{ 4 }-\frac { 3 }{ 4 } \\ \\ =\frac { 3 }{ 4 } \left( { 3 }^{ 4 }-1 \right)$

$\\ \\ =\frac { 3 }{ 4 } \cdot 80\\ \\ =60$

60 units squared.

Answer from: Quest

Answer: choice c and choice d

=====================================

explanation

the graph has a slope of 1/3 because each time you move up 1, you go to the right 3. so, slope = rise/run = 1/3. we can cross choice a off the list since we want a slope of 3, rather than 1/3.

the first table shown has a slope of -3. this is because the rise is -6 and the run is 2, so slope = rise/run = -6/2 = -3. note how x increases by 2 each time (run) and y drops by 6 each time (rise). choice b can be eliminated.

the second table has a slope of 3. each time x goes up by 1, the y values increase by 3. slope = rise/run = 3/1 = 3. alternatively you can use the slope formula to prove that the slope is 3 here. use any two rows of values to generate the two points. choice c is one of the answers

the third table also has a slope of 3. let's use the slope formula to prove as such. i'm going to use the first two rows of the table. so i'll use the two points (2,1) and (5,10)

m = slope

m = (y2-y1)/(x2-x1)

m = (10-1)/(5-2)

m = 9/3

m = 3 so we get a slope of 3

choice d is the other answer

Which functions have an additive rate of change of 3? check all that apply

Answer from: Quest

(1, –1), (16, –4), (36, –6)

step-by-step explanation:

given that the function is

$f(x) =\sqrt{x}$

this can also be written as

$y^{2} =x$

we have to graph this function and check which list of points lie on the graph

see attachment for graph:

we find that the points lying are (1,-1) (16,-4) and (36,-6)

since square root is there we find that x can only be posiitve but y can be since square root both positive and negative.

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Ex 3.7
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