Practice for Exam Four

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Practice for Exam Four.

1 :

Compute the following sum: $\sum \limits _{k=\sage {p1indexStart}}^{\sage {p1indexEnd}} \sage {p1f1(x=k)} = \answer {\sage {p1ans}}$

Remember that the index ( $k$ in this case) starts at the start value ( $\sage {p1indexStart}$ ) and ends at the end value ( $\sage {p1indexEnd}$ ) and includes both values, so you should be adding a total of $\sage {p1totalTerms}$ terms being added together.

Also don’t forget that Xronos doesn’t need you to simplify your answers here, so don’t worry about making everything “nice”. This is just to make sure you know how the notation works!

2 :

Compute the indefinite integral of $f(x) = \sage {p3f1}$ : $\int \sage {p3f1}dx = \answer [validator=sameDerivative]{\sage {p3ans}}$

Remember that you want the antiderivative of the given function, but not just one antiderivative but all of them. So after you find an antiderivative, you need to include the $+C$ (case-sensitive!) to denote that you are including the entire class of antiderivatives.

3 :

Use limits to find a perfect approximation for the area under the curve of the function $f(x) = \sage {p4f1}$ from $x=\sage {p4start}$ to $x=\sage {p4end}$ . $\answer {\sage {p4ans}}$ .

Remember you can factor out any constants from the function all the way outside of the sum, which will help with the limits. You want to make sure to break the sum apart as much as you can and use the handy identities:

(a): $\sum \limits _{k=1}^N 1 = N$
(b): $\sum \limits _{k=1}^N k = \frac {N(N+1)}{2}$
(c): $\sum \limits _{k=1}^N N^2 = \frac {N(N+1)(2N+1)}{6}$ .

This will help you get rid of the $\Sigma$ part and get to a “closed form” where you can actually evaluate the limit!

4 :

Compute the indefinite integral of $f(x) = \sage {p5fdisp}$ : $\int \sage {p5fdisp}dx = \answer [validator=sameDerivative]{\sage {p5f1}}$

Don’t forget you can always factor out any constants that are multiplying the whole integrand, to make your life a little easier while you tackle the problem. Then look for functions inside other functions that have a derivative multiplying everything outside. In this case, try doing a substitution like $u = \sage {p5finside}$ and see if that helps. Don’t forget to calculate $du$ , and as always, don’t forget your friend the $+C$ !

5 : Find an antiderivative of $f(x) = \sage {p6f1}$ : $F(x) = \answer [validator=sameDerivative]{\sage {p6ans}}$ .

Remember that the antiderivative is not the same as the indefinite integral, and here you want to reverse the polynomial derivative process.

6 : Find an antiderivative of $f(x) = \sage {p7f1}$ : $F(x) = \answer [validator=sameDerivative]{\sage {p7ans}}$ .

Remember that the antiderivative is not the same as the indefinite integral, and here you want to reverse the exponential derivative process.

7 : Find an antiderivative of $f(x) = \sage {p8f1}$ : $F(x) = \answer [validator=sameDerivative]{\sage {p8ans}}$ .

Remember that the antiderivative is not the same as the indefinite integral, and here you want to reverse the polynomial derivative process.

8 : Consider the following: $\int \limits _{\sage {p9c1}}^{\sage {p9c2}} f(x) dx$ What does each of these parts of the notation mean?

The $\int \limits _{\sage {p9c1}}^{\sage {p9c2}}$ means:
We want the indefinite integral of $f(x)$ . We want the derivative of $f(x)$ . We want the unsigned area between the $x$ axis and the graph of $f(x)$ from $x=\sage {p9c1}$ to $x=\sage {p9c2}$ . We want the unsigned area between the $x$ axis and the graph of $f(x)$ from $x=\sage {p9c2}$ to $x=\sage {p9c1}$ . We want the signed area between the $x$ axis and the graph of $f(x)$ from $x=\sage {p9c1}$ to $x=\sage {p9c2}$ . We want the signed area between the $x$ axis and the graph of $f(x)$ from $x=\sage {p9c2}$ to $x=\sage {p9c1}$ . We want to integrate with respect to the $x$ variable. We want to integrate with respect to the $d$ variable. The integrand, which is the function whose graph we use in the integration process.
The $dx$ means:
We want the indefinite integral of $f(x)$ . We want the derivative of $f(x)$ . We want the unsigned area between the $x$ axis and the graph of $f(x)$ from $x=\sage {p9c1}$ to $x=\sage {p9c2}$ . We want the unsigned area between the $x$ axis and the graph of $f(x)$ from $x=\sage {p9c2}$ to $x=\sage {p9c1}$ . We want the signed area between the $x$ axis and the graph of $f(x)$ from $x=\sage {p9c1}$ to $x=\sage {p9c2}$ . We want the signed area between the $x$ axis and the graph of $f(x)$ from $x=\sage {p9c2}$ to $x=\sage {p9c1}$ . We want to integrate with respect to the $x$ variable. We want to integrate with respect to the $d$ variable. The integrand, which is the function whose graph we use in the integration process.
The $f(x)$ is:
We want the indefinite integral of $f(x)$ . We want the derivative of $f(x)$ . We want the unsigned area between the $x$ axis and the graph of $f(x)$ from $x=\sage {p9c1}$ to $x=\sage {p9c2}$ . We want the unsigned area between the $x$ axis and the graph of $f(x)$ from $x=\sage {p9c2}$ to $x=\sage {p9c1}$ . We want the signed area between the $x$ axis and the graph of $f(x)$ from $x=\sage {p9c1}$ to $x=\sage {p9c2}$ . We want the signed area between the $x$ axis and the graph of $f(x)$ from $x=\sage {p9c2}$ to $x=\sage {p9c1}$ . We want to integrate with respect to the $x$ variable. We want to integrate with respect to the $d$ variable. The integrand, which is the function whose graph we use in the integration process.

Remember that, in order for a limit to exist, both the left and right limits must exist and be equal.

9 :

Use $\sage {p10RecNum}$ rectangles to approximate the area under the curve of the function $f(x) = \sage {p10f1}$ from $x=\sage {p10start}$ to $x=\sage {p10end}$ using...

Right Endpoint Approximation Method: $\answer {\sage {p10ansREP}}$
Left Endpoint Approximation Method: $\answer {\sage {p10ansLEP}}$
Midpoint Approximation Method: $\answer {\sage {p10ansMP}}$

Remember to use the width and height of the rectangles to make your formula. In this case you should have a rectangle width of $\Delta x = \frac {(\text { end point }) - (\text { start point })}{\text {Number of Rectangles}} = \frac {(\sage {p10end})-(\sage {p10start})}{\sage {p10RecNum}} = \sage {p10width}$ .

Also don’t forget that Xronos doesn’t need you to simplify your answers here, so don’t worry about making everything “nice”. You will eventually need to simplify things in a later segment, so it’s a good idea to practice, but if you are struggling try entering in the unsimplified answer to see if you are getting the right idea but making an arithmetic error when simplifying.

10 : Use the fundamental theorem of calculus to compute the following definite integral: $\int \limits _{\sage {p11left}}^{\sage {p11right}} \sage {p11f1}dx = \answer {\sage {p11ans}}$

The fundamental theorem of calculus lets you use the antiderivative of the integrand to compute the area under the curve, no more Riemann Approximation needed!