Week 8 Activity

Section 8.4 Week 8 Activity

Subsection 8.4.1 Sigma Notation

Activity 8.4.1.

Write this sum in ordinary notation.

\begin{equation*} \sum_{i=2}^9 i^2 \end{equation*}

Solution.

Starting at 2, I replace \(i\) with all the numbers up to 9 and then add all these results together.

\begin{align*} \amp \sum_{i=2}^9 i^2 = 2^2 + 3^2 + 4^2 + 5^2 + 6^2 + 7^2 + 8^2 + 9^2 \\ \amp = 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 \end{align*}

Activity 8.4.2.

Write this sum in ordinary notation.

\begin{equation*} \sum_{i=-3}^{10} (6i-4) \end{equation*}

Solution.

Starting at \(-3\text{,}\) I replace \(i\) with all the numbers up to 10 and then add all these results together.

\begin{align*} \amp \sum_{i=-3}^{10} (6i-4) = (6(-3) - 4) + (6(-2) - 4) + (6(-1) - 4) + (6(0) - 4)\\ \amp + (6(1) - 4) + (6(2) - 4) + (6(3) - 4) + (6(4) - 4) + (6(5) - 4) \\ \amp + (6(6) - 4) + (6(7) - 4) + (6(8) - 4) + (6(9) - 4) + (6(10) - 4) \\ \amp = -22 + (-16) + (-10) + (-4) + 2 + 8 + 14 + 20 + 26 + 32\\ \amp + 38 + 44 + 50 + 56 \end{align*}

Activity 8.4.3.

Write this sum in ordinary notation.

\begin{equation*} \sum_{i=0}^{15} 2^i \end{equation*}

Solution.

Starting with 0, I replace \(i\) with all the numbers up to 15 and then add all the resulting numbers together.

\begin{align*} \amp \sum_{i=0}^{15} 2^i = 2^0 + 2^1 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 \\ \amp + 2^9 + 2^{10} + 2^{11} + 2^{12} + 2^{13} + 2^{14} + 2^{15} \\ \amp = 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 + 512 + 1024\\ \amp + 2048 + 4096 + 8192 + 16384 + 32768 \end{align*}

Activity 8.4.4.

Write this sum in ordinary notation.

\begin{equation*} \sum_{i=3}^7 (i^2 + 5i - 3^i) \end{equation*}

Solution.

Starting with 3, I replace \(i\) will all the number until 7. Then I add up all the resulting numbers.

\begin{align*} \amp \sum_{i=3}^7 (i^2 + 5i - 3^i) = (3^2 + 5(3) - 3^3) + (4^2 + 5(4) - 3^4) + (5^2 + 5(5) - 3^5)\\ \amp + (6^2 + 5(6) - 3^6) + (7^2 + 5(7) - 3^7) \\ \amp = (9 + 15 - 27) + (16 + 20 - 81) + (25 + 25 - 243)\\ \amp + (36 + 30 - 729) + (49 + 35 - 2187)\\ \amp = (-3) + (-45) + (-193) + (-663) + (-2103) \end{align*}

Activity 8.4.5.

Write this sum in sigma notation.

\begin{equation*} 1 + 4 + 7 + 10 + 13 + 16 + 19 \end{equation*}

Solution.

I need a pattern. This is an arithmetic sequence, where each step is adding 3. I start with a value of 1. Using this, I can start at \(i=0\) and describe this pattern as \(1 + 3i\text{.}\) The value of \(i\) that gives 19 is \(i=6\text{,}\) so the sum will end at 6.

\begin{equation*} 1 + 4 + 7 + 10 + 13 + 16 + 19 = \sum_{i=0}^6 (1+3i) \end{equation*}

Activity 8.4.6.

Write this sum in sigma notation.

\begin{equation*} 3 + 6 + 12 + 24 + 48 + 96 \end{equation*}

Solution.

I need a pattern. This is a geometric sequence, where each steps multiplies by 2. I start with a value of 3. Using this, I can start at \(i=0\) and describe this pattern as \(3(2^i)\text{.}\) The value of \(i\) that gives 96 is \(i = 5\text{,}\) so the sum will end at 6.

\begin{equation*} 3 + 6 + 12 + 24 + 48 + 96 = \sum_{i=0}^5 3(2^i) \end{equation*}

Activity 8.4.7.

Write this sum in sigma notation.

\begin{equation*} 16 + 14 + 12 + 10 + 8 + 6 + 4 \end{equation*}

Solution.

I need a pattern. I have a couple options. I could keep this in ths order is it written in, start from 16, and subtract 2 in each step. I could also change the order and write this as an arithmetic progression starting at 4. I’ll chose the first option, starting with \(i=0\) and writing \(16 -2i\) for the pattern. When \(i=6\text{,}\) I get the last term, so the sum will end at 6.

\begin{equation*} 16 + 14 + 12 + 10 + 8 + 6 + 4 = \sum_{i=0}^6 (16-2i) \end{equation*}

Activity 8.4.8.

Use linearity to split this into several sums with constants removed.

\begin{equation*} \sum_{i=0}^{44} (5i^3 - 2i^2 - 19i + 19) \end{equation*}

Solution.

Linearity splits up the sum over addition and subtraction and allows me to pull out the constants. In each piece, the index and bounds of the sum stay the same.

\begin{equation*} \sum_{i=0}^{44} (5i^3 - 2i^2 - 19i + 19) = 5 \sum_{i=0}^{44} i^3 - 2 \sum_{i=0}^{44} i^2 - 19 \sum_{i=0}^{44} i + 19 \sum_{i=0}^{44} 1 \end{equation*}

Activity 8.4.9.

Use linearity to split this into several sums with constants removed.

\begin{equation*} \sum_{i=4}^{104} (-12i^5 + i^3 - 71i^2 + 4) \end{equation*}

Solution.

Linearity splits up the sum over addition and subtraction and allows me to pull out the constants. In each piece, the index and bounds of the sum stay the same.

\begin{equation*} \sum_{i=4}^{104} (-12i^5 + i^3 - 71i^2 + 4) = -12 \sum_{i=4}^{104} i^5 + \sum_{i=4}^{104} i^3 - 71 \sum_{i=4}^{104} i^2 + 4 \sum_{i=4}^{104} 1 \end{equation*}

Activity 8.4.10.

Use linearity to combine this into one sum.

\begin{equation*} 4 \sum_{i=0}^{10} i^3 - 15 \sum_{i=0}^{10} i^2 - 7 \sum_{i=0}^{10} i + 9 \sum_{i=0}^{10} i \end{equation*}

Solution.

The indices and bounds of all the sums are the same, so I can combine them into one sum by taking the constants inside the sums and writing the additions and subtractions inside the sum. Notice there are two \(i\) terms, which I can combine into one.

\begin{align*} 4 \sum_{i=0}^{10} i^3 - 15 \sum_{i=0}^{10} i^2 - 7 \sum_{i=0}^{10} i + 9 \sum_{i=0}^{10} i \amp = \sum_{i=0}^{10} (4i^3 - 15i^2 - 7i + 9i) \\ \amp = \sum_{i=0}^{10} (4i^3 - 15i^2 + 2i) \end{align*}

Activity 8.4.11.

Use linearity to combine this into one sum.

\begin{equation*} 19 \sum_{i=4}^{17} i^3 - \sum_{i=4}^{17} i^3 + 2 \sum_{i=4}^{17} i^2 - 36 \sum_{i=4}^{17} i \end{equation*}

Solution.

\begin{align*} 19 \sum_{i=4}^{17} i^3 - \sum_{i=4}^{17} i^3 + 2 \sum_{i=4}^{17} i^2 - 36 \sum_{i=4}^{17} i \amp = \sum_{i=4}^{17} (19i^3 - i^3 + 2i^2 - 36 i) \\ \amp = \sum_{i=4}^{17} (18i^3 + 2i^2 - 36 i) \end{align*}

Subsection 8.4.2 Riemann Integral

Activity 8.4.12.

This is a multi-part activity to solve the integral

\begin{equation*} \int_{-1}^1 (x^2 + 2x + 2)dx \end{equation*}

by definition using the Reimann definition of the integral. First, write the definition of the integral and replace the general pieces with the specifics of this function.

Solution.

Here is the general form.

\begin{equation*} \int_a^b f(x) dx = \lim_{n \rightarrow \infty} \sum_{k=1}^n f(x_k^*) \left( \frac{b-a}{n} \right) \end{equation*}

In this situation, \(a = -1\text{,}\) \(b = 1\) and \(f(x) = (x^2 + 2x +2)\text{.}\) I put these into the general form.

\begin{align*} \int_{-1}^1 (x^2 + 2x + 2) dx \amp = \lim_{n \rightarrow \infty} \sum_{k=1}^n \left( (x_k^*)^2 + 2x_k^* + 2 \right) \left( \frac{1-(-1)}{n} \right) \\ \amp = \lim_{n \rightarrow \infty} \sum_{k=1}^n \left( (x_k^*)^2 + 2x_k^* + 2 \right) \left( \frac{2}{n} \right) \end{align*}

Activity 8.4.13.

When write an expression for \(x_k^*\) and use it in the expression.

Solution.

There is a general form for \(x_k^*\) that I have been using in the examples.

\begin{equation*} x_k^* = a + k \frac{b-a}{n} \end{equation*}

Now I put in the values for this particular integral.

\begin{equation*} x_k^* = -1 + k \frac{1-(-1)}{n} = -1 + \frac{2k}{n} \end{equation*}

Then I can return to the form from the previous section and replace \(x_k^*\) with this new expression. I’ll proceed to simplify the polynomial after I’ve substituted \(x_k^*\text{.}\)

\begin{align*} \int_{-1}^1 (x^2+2+2) dx \amp = \lim_{n \rightarrow \infty} \sum_{k=1}^n \left( \left( -1 + \frac{2k}{n} \right)^2 + 2 \left( -1 + \frac{2k}{n} \right) + 2 \right) \left( \frac{2}{n} \right) \\ \amp = \lim_{n \rightarrow \infty} \sum_{k=1}^n \left( 1 - \frac{4k}{n} + \frac{4k^2}{n^2} + -2 - \frac{4k}{n} + 2 \right) \left( \frac{2}{n} \right) \\ \amp = \lim_{n \rightarrow \infty} \sum_{k=1}^n \left( 1 + \frac{4k^2}{n^2} \right) \left( \frac{2}{n} \right) \\ \amp = \lim_{n \rightarrow \infty} \sum_{k=1}^n \left( \frac{2}{n} + \frac{8k^2}{n^3} \right) \end{align*}

Activity 8.4.14.

Then rearrange and split up the sum using linearity.

Solution.

There are two pieces of this sum from the end of the previous part. I split this into two sums and pull out the constants. Note that \(k\) is the index of the sum, so anything that doesn’t involve \(k\) is a constant. That means that \(n\) is a constant as far as the sum is concerned and can be brought outside.

\begin{align*} \int_{-1}^1 (x^2+2x+2) dx \amp = \lim_{n \rightarrow \infty} \sum_{k=1}^n \left( \frac{2}{n} + \frac{8k^2}{n^3} \right) \\ \amp = \lim_{n \rightarrow \infty} \left( \sum_{k=1}^n \frac{2}{n} + \sum_{k=1}^n \frac{8k^2}{n^3} \right)\\ \amp = \lim_{n \rightarrow \infty} \left( \frac{2}{n} \sum_{k=1}^n 1+ \frac{8}{n^3} \sum_{k=1}^n k^2 \right) \end{align*}

Activity 8.4.15.

Then replace the remaining sums with their values according to the standard sum formulas.

Solution.

From the previous step, there are sums with terms 1 and \(k^2\text{.}\) These are standard sums which can be evaluated. After the replacement, I can simplify the expression a bit.

\begin{align*} \amp \lim_{n \rightarrow \infty} \left( \frac{2}{n} \sum_{k=1}^n 1 + \frac{8}{n^3} \sum_{k=1}^n k^2 \right)\\ \amp = \lim_{n \rightarrow \infty} \left( \frac{2}{n} n + \frac{8}{n^3} \frac{n(n+1)(2n+1)}{6} \right)\\ \amp = \lim_{n \rightarrow \infty} \left( 2 + \frac{8n(n+1)(2n+1)}{6n^3} \right)\\ \amp = \lim_{n \rightarrow \infty} \left( 2 + \frac{8n(2n^2 + 3n + 1)}{6n^3} \right)\\ \amp = \lim_{n \rightarrow \infty} \left( 2 + \frac{16n^3 + 24n^2 + 8n)}{6n^3} \right) \end{align*}

Activity 8.4.16.

Finally, use asymptotic analysis to solve the limit. Write the final value of the integral.

Solution.

The constant term 2 remains unchanged in the limit. The other term is a fraction with \(n^3\) in the numerator and denominator. Since these have the same asymptotic order, asymptotic analysis tells that the limit is the ratio of the leading coefficients.

\begin{equation*} \lim_{n \rightarrow \infty} \left( 2 + \frac{16n^3 + 24n^2 + 8n)}{6n^3} \right) = 2 + \frac{16}{6} = 2 + \frac{8}{3} = \frac{14}{3} \end{equation*}

This completes the integral. The area under the curve described by the original problem is \(\frac{14}{3}\) units squared.

Subsection 8.4.3 Fundamental Theorem of Calculus

Activity 8.4.17.

Write the following antiderivative by using known derivatives or doing the derivative rules backwards.

\begin{equation*} \int -\sin x dx \end{equation*}

Solution.

I look to the tables of derivatives. \(-\sin x\) shows up as the derivative of \(\cos x\text{,}\) so I go backwards to that original function.

\begin{equation*} \int -\sin x dx = \cos x + c \end{equation*}

Activity 8.4.18.

Write the following antiderivative by using known derivatives or doing the derivative rules backwards.

\begin{equation*} \int \frac{1}{x} dx \end{equation*}

Solution.

I look to the tables of derivatives. If I differentiate \(\ln x\text{,}\) I get \(\frac{1}{x}\text{,}\) so I go backwards to that original function.

\begin{equation*} \int \frac{1}{x} dx = \ln x + c \end{equation*}

If I look at the tables of integrals, I can actually do a bit better than this. The right side of the above equation is only defined for \(x > 0\text{,}\) but the left side is defined for all \(x \neq 0\text{.}\) There is an antiderivative that matches the full domain.

\begin{equation*} \int \frac{1}{x} dx = \ln |x| + c \end{equation*}

Activity 8.4.19.

Write the following antiderivative by using known derivatives or doing the derivative rules backwards. You can use the linearity of the integral.

\begin{equation*} \int (4 \cos x + 6 \sin x) dx \end{equation*}

Solution.

I can use linearity to split this into two integrals.

\begin{equation*} \int (4 \cos x + 6 \sin x) dx = 4 \int \cos x dx + 6 \int \sin x dx \end{equation*}

Then I look at the derivative tables and work backwards to get the antiderivatives of sine and cosine.

\begin{equation*} = 4 \int \cos x dx + 6 \int \sin x dx = 4 \sin x - 6 \cos x + c \end{equation*}

Activity 8.4.20.

Write the following antiderivative by using known derivatives or doing the derivative rules backwards. You can use the linearity of the integral.

\begin{equation*} \int (x^3-e^x)dx \end{equation*}

Solution.

I can use linearity to split this into two integrals.

\begin{equation*} \int (x^3-e^x)dx = \int x^3 dx - \int e^x dx \end{equation*}

For the first integral, I have to do the power rule backwards. This increases the exponent and divides by the new exponent. For the second integral, since the exponential function is its own derivative, it is also its own antiderivative.

\begin{equation*} \int x^3 dx - \int e^x dx = \frac{x^4}{4} - e^x + c \end{equation*}

Activity 8.4.21.

Write the following antiderivative by using known derivatives or doing the derivative rules backwards. However, don’t do the chain rule backwards; instead, use what you know about the chain rule to make a reasonable guess for the antiderivative.

\begin{equation*} \int e^{4x} dx \end{equation*}

Solution.

If I were to differentiate this, it would be a chain rule derivative. The outside function would remain unchanged, but the inside derivative is \(\frac{d}{dx} 4x = 4\text{,}\) so I would have to multiply by 4. Going backwards, then, I should do the opposite and divide by 4.

\begin{equation*} \int e^{4x} dx = \frac{1}{4} e^{4x} + c \end{equation*}

Activity 8.4.22.

Write the following antiderivative by using known derivatives or doing the derivative rules backwards.

\begin{equation*} \int_0^{\pi} \sin x dx \end{equation*}

Solution.

The antiderivative of \(\sin x\) is \((-\cos x)\text{.}\) For definite integrals, after I find the antiderivative, I evaluate it on the bounds. (I don’t need the constant of integral in definite integrals, since it would just cancel off when I evaluate on the bounds.)

\begin{equation*} \int_0^{\pi} \sin x dx = -\cos x \Big|_0^{\pi} = -\cos (\pi) - (- \cos (0)) = -(-1) - (-1) = 2 \end{equation*}

Activity 8.4.23.

Write the following antiderivative by using known derivatives or doing the derivative rules backwards.

\begin{equation*} \int_1^{\ln 4} e^x dx \end{equation*}

Solution.

The antiderivative of the exponential it itself. I need to evaluate on the bounds. Remember that the logarithm and the exponential are inverse operations, so they remove each other.

\begin{equation*} \int_1^{\ln 4} e^x dx = e^x \Big|_1^{\ln 4} = e^{\ln 4} - e^{1} = 4 - e \end{equation*}

Activity 8.4.24.

Write the following antiderivative by using known derivatives or doing the derivative rules backwards.

\begin{equation*} \int_1^5 (2x-7) dx \end{equation*}

Solution.

I use linearity to split up the integral.

\begin{equation*} \int_1^5 (2x-7) dx = \int_1^5 2x dx - \int_1^5 7 dx \end{equation*}

Then I use the power rule backwards for the first integral. In the second, the antiderivative of a constant is just the variable itself.

\begin{align*} \int_1^5 2x dx - \int_1^5 7 dx \amp = \frac{2x^2}{2} \Big|_1^5 - 7x \Big|_1^5 \\ \amp = 5^2 - 1^2 - (7(5) - 7) = 25 - 1 - 35 + 7 = -4 \end{align*}

Subsection 8.4.4 Conceptual Review Questions

What is sigma notation and why is it used?
Why are sums linear?
What is a definite integral?
How does the definition of the integral use a limit?
How does the integral because exact, instead of just an approximation, after the limit is calculated?
What is an antiderivative?
How are the derivative and integral tables mirrors of each other?
What is the difference between a definite and indefinite integral?

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