6.3: Homework- Numeric Integration Techniques
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Given the picture of \(f(x)\), find the following definite integrals.
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\(\int_{-2}^2 f(x)dx\).
\(2\)ans
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\(\int_{2}^7 f(x)dx\).
\(8\)ans
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\(\int_0^6 f(x)dx\).
\(8\)ans
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\(\int_4^7 f(x)dx\).
\(0\)ans
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\(\int_{-2}^7 f(x)dx\).
\(10\)ans
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\(\int_{3}^3 f(x)dx\) (what do you think this even means?)
\(0\) — You can think of this as an infinitely thin rectangle.ans
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\(\int_{-1}^{0} f(x)dx\).
\(1\)ans
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\(\int_{0}^{1.5} f(x)dx\).
ans
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\(\int_{-1.5}^{4.5} f(x)dx\)
\(10.5\)ans
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\(\int_{-2}^2 f(x)dx\).
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Remember those problems from homework 1 where we started with a velocity graph and then drew the position graph? Let’s try that again.
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Draw the position graph corresponding to this velocity graph next to the graph above.
ans
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Find the following three integrals based on the original graph \(v(t)\):
\(\int_0^3 v(t)dt, \quad \int_0^6 v(t)dt, \quad \int_0^9 v(t)dt.\)
\(60\), \(30\), \(30\)ans -
Your answers for parts (a) and (b) should look the same in some way. How do they look the same, and why did it work out this way?
The answers are the key \(y\)-values from the graph in part (a). This is the same, since in both cases you are doing the same thing: takeing \(\Delta x \cdot y\) (or \(\Delta t \cdot \text{velocity}\) and adding it up as you go.ans
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Draw the position graph corresponding to this velocity graph next to the graph above.
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Using \(6\) rectangles and using the left rectangle rule, estimate the area under the curve of the following functions from \(x = 0\) to \(x = 3\).
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\(e(x) = 2 - x\).
\(\approx 2.25\)ans
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\(f(x) = -\frac{1}{9} x^2 + 1\)
\(\approx 2.24\)ans
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\(g(x) = e^x\)
\(\approx 14.71\)ans
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\(e(x) = 2 - x\).
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For \(e(x)\) in problem 3, find the exact area using the formula for the area of a triangle.
\(1.5\)ans
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For \(f(x)\) in problem 3:
- If you haven’t done so already, sketch a picture of the graph as well as the rectangles you used to approximate the area.
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Is your approximation from problem 3 an
over-estimate
or an
under-estimate
? How do you know?
Overestimate, since it looks like the rectangles cover too much area.ans
- Watch the Khan Academy video on the Trapezoidal method of finding the area under the curve .
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Using \(6\) trapezoids that match the height of the graph, estimate the area under the curve of the following functions from \(x = 0\) to \(x = 3\).
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\(e(x) = 2 - x\).
\(1.5\)ans
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\(f(x) = -\frac{1}{9} x^2 + 1\)
\(1.99\)ans
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\(g(x) = e^x\)
\(19.48\)ans
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\(e(x) = 2 - x\).
- Watch this Khan Academy like video on the midpoint formula: click here
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Using \(6\) rectangles, use the midpoint rule to approximate the area under the curve from \(x = 0\) to \(x = 3\).
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\(e(x) = 2 - x\).
\(1.5\)ans
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\(f(x) = -\frac{1}{9} x^2 + 1\)
\(2.03\)ans
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\(g(x) = e^x\)
\(18.31\)ans
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\(e(x) = 2 - x\).
- I couldn’t find a Khan Academy explanation of Simpson’s rule, so here is another video by Patrick at Just Math Tutoring .
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Use Simpson’s rule to approximate the area under the curve. Use \(6\) intervals from \(x = 0\) to \(x = 3\).
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\(e(x) = 2 - x\).
\(1.5\)ans
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\(f(x) = -\frac{1}{9} x^2 + 1\)
\(2\)ans
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\(g(x) = e^x\)
\(\approx 19.09\)ans
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\(e(x) = 2 - x\).
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Note that the actual answers for the area under the curve from \(x = 0\) to \(x = 3\) are
- \(1.5\) for \(e(x) = 2-x\).
- \(2\) for \(f(x) = -\frac{1}{9} x^2 + 1\)
- \(19.086\) for \(g(x) = e^x\).
Given these answers, rate the following rules from most accurate to least accurate based on the answers from this homework: left rectangle rule, trapezoid rule, midpoint rule, Simpson’s rule.