4.5.1 AP Calculus Exam .... area of piece wise funcion

In summary, Yahoo had an answer to this but it's never in Latex so I couldn't understand how they got $\dfrac{7}{2}$.
  • #1
karush
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ok well it isn't just adding the areas of 2 functions but is $xf(x)$ as an integrand

Yahoo had an answer to this but its never in Latex so I couldn't understand how they got $\dfrac{7}{2}$
 
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  • #2
karush said:
ok well it isn't just adding the areas of 2 functions but is $xf(x)$ as an integrand

Yahoo had an answer to this but its never in Latex so I couldn't understand how they got $\dfrac{7}{2}$

What have you tried? Why can't you split it up at x = 0. Is xf(x) ever negative on that interval?
 
  • #3
Since f(x)= x if [tex]x\le 0[/tex] and f(x)= x+ 1 if x> 0, [tex]xf(x)= x^2[/tex] if [tex]x\le 0[/tex] and [tex]xf(x)= x^2+ x[/tex] if x> 0. So [tex]\int_{-2}^1 xf(x)dx= \int_{-2}^0 x^2 dx+ \int_0^1 x^2+ x dx[/tex].

[tex]\int_{-2}^0 x^2 dx= \left[\frac{1}{3}x^3\right]_{-2}^0= \frac{1}{3}(0- (-8))= \frac{8}{3}[/tex].

[tex]\int_0^1 x^2+ x dx= \left[\frac{1}{3}x^3+ \frac{1}{2}x^2\right]_0^1= \frac{1}{3}+ \frac{1}{2}= \frac{5}{6}[/tex].

[tex]\int_{-2}^1 xf(x) dx= \frac{8}{3}+ \frac{5}{6}= \frac{16}{6}+ \frac{5}{6}= \frac{21}{6}= \frac{7}{2}[/tex].
 
  • #4
I thot of doing that but didn't think it was correct🤔

These exam questions you got to be very careful they are very subtle
 
  • #5
HallsofIvy said:
Since f(x)= x if [tex]x\le 0[/tex] and f(x)= x+ 1 if x> 0, [tex]xf(x)= x^2[/tex] if [tex]x\le 0[/tex] and [tex]xf(x)= x^2+ x[/tex] if x> 0. So [tex]\int_{-2}^1 xf(x)dx= \int_{-2}^0 x^2 dx+ \int_0^1 x^2+ x dx[/tex].

[tex]\int_{-2}^0 x^2 dx= \left[\frac{1}{3}x^3\right]_{-2}^0= \frac{1}{3}(0- (-8))= \frac{8}{3}[/tex].

[tex]\int_0^1 x^2+ x dx= \left[\frac{1}{3}x^3+ \frac{1}{2}x^2\right]_0^1= \frac{1}{3}+ \frac{1}{2}= \frac{5}{6}[/tex].

[tex]\int_{-2}^1 xf(x) dx= \frac{8}{3}+ \frac{5}{6}= \frac{16}{6}+ \frac{5}{6}= \frac{21}{6}= \frac{7}{2}[/tex].
mahalo
 
  • #6
karush said:
I thot of doing that but didn't think it was correct🤔

These exam questions you got to be very careful they are very subtle

This one is not anywhere near "subtle". It wants to know if you know that you can partition the Domain of the integral. This is a very important, fundamental principle.
 

FAQ: 4.5.1 AP Calculus Exam .... area of piece wise funcion

1. What is a piecewise function?

A piecewise function is a function that is defined by different equations over different intervals of its domain. This means that the function may have different rules or formulas for different parts of its input values.

2. How do you find the area of a piecewise function?

To find the area of a piecewise function, you need to break it down into smaller parts that can be calculated separately. Then, you can use the appropriate formula (such as the area under a curve or the area of a rectangle) for each part and add them together to get the total area.

3. What is the difference between a piecewise function and a continuous function?

A continuous function is a function that can be drawn without lifting your pencil from the paper. This means that there are no gaps or holes in the graph. A piecewise function, on the other hand, may have breaks or discontinuities in its graph where the different equations meet.

4. Can a piecewise function have more than two pieces?

Yes, a piecewise function can have as many pieces as needed. There is no limit to the number of equations that can make up a piecewise function, as long as each equation corresponds to a different interval of the function's domain.

5. How do you determine the domain and range of a piecewise function?

The domain and range of a piecewise function are determined by the individual domains and ranges of each equation that makes up the function. You need to consider the restrictions and limitations of each equation and combine them to determine the overall domain and range of the piecewise function.

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