Difference between revisions of "009B Sample Final 1, Problem 6"
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!Step 1: | !Step 1: | ||
|- | |- | ||
| − | | | + | |First, we write <math>\int_1^4 \frac{dx}{\sqrt{4-x}}=\lim_{a\rightarrow 4} \int_1^a\frac{dx}{\sqrt{4-x}}</math>. |
|- | |- | ||
| − | | | + | |Now, we proceed by <math>u</math>-substitution. We let <math>u=4-x</math>. Then, <math>du=-dx</math>. |
|- | |- | ||
| − | | | + | |Since the integral is a definite integral, we need to change the bounds of integration. |
| + | |- | ||
| + | |Plugging in our values into the equation <math>u=4-x</math>, we get <math>u_1=4-1=3</math> and <math>u_2=4-a</math>. | ||
| + | |- | ||
| + | |Thus, the integral becomes | ||
| + | |- | ||
| + | |<math>\int_1^4 \frac{dx}{\sqrt{4-x}}=\lim_{a\rightarrow 4} \int_3^{4-a}\frac{-du}{\sqrt{u}}</math>. | ||
|} | |} | ||
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!Step 2: | !Step 2: | ||
|- | |- | ||
| − | | | + | |We integrate to get |
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| − | |||
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| − | |||
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|- | |- | ||
| − | | | + | |<math>\int_1^4 \frac{dx}{\sqrt{4-x}}=\lim_{a\rightarrow 4} \left.-2u^{\frac{1}{2}}\right|_{3}^{4-a}=\lim_{a\rightarrow 4}-2\sqrt{4-a}+2\sqrt{3}=2\sqrt{3}</math> |
|} | |} | ||
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|'''(a)''' <math>1</math> | |'''(a)''' <math>1</math> | ||
|- | |- | ||
| − | |'''(b)''' | + | |'''(b)''' <math>2\sqrt{3}</math> |
|} | |} | ||
[[009B_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] | [[009B_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] | ||
Revision as of 15:36, 2 February 2016
Evaluate the improper integrals:
- a)
- b) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_1^4 \frac{dx}{\sqrt{4-x}}}
| Foundations: |
|---|
| Review integration by parts |
Solution:
(a)
| Step 1: |
|---|
| First, we write Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_0^{\infty} xe^{-x}~dx=\lim_{a\rightarrow \infty} \int_0^a xe^{-x}~dx} . |
| Now, we proceed using integration by parts. Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u=x} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle dv=e^{-x}dx} . Then, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle du=dx} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v=-e^{-x}} . |
| Thus, the integral becomes |
| Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_0^{\infty} xe^{-x}~dx=\lim_{a\rightarrow \infty} \left.-xe^{-x}\right|_0^a-\int_0^a-e^{-x}dx} |
| Step 2: |
|---|
| For the remaining integral, we need to use Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u} -substitution. Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u=-x} . Then, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle du=-dx} . |
| Since the integral is a definite integral, we need to change the bounds of integration. |
| Plugging in our values into the equation Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u=-x} , we get Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u_1=0} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u_2=-a} . |
| Thus, the integral becomes |
| Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_0^{\infty} xe^{-x}~dx=\lim_{a\rightarrow \infty} \left.-xe^{-x}\right|_0^a-\int_0^{-a}e^{u}du=\lim_{a\rightarrow \infty} \left.-xe^{-x}\right|_0^a-\left.e^{u}\right|_0^{-a}=\lim_{a\rightarrow \infty} -ae^{-a}-(e^{-a}-1)} |
| Step 3: |
|---|
| Now, we evaluate to get |
| Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_0^{\infty} xe^{-x}~dx=\lim_{a\rightarrow \infty} -ae^{-a}-(e^{-a}-1)=\lim_{a\rightarrow \infty} \frac{-a}{e^a}-\frac{1}{e^a}+1=\lim_{a\rightarrow \infty} \frac{-a-1}{e^a}+1} . |
| Using L'Hopital's Rule, we get |
| Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_0^{\infty} xe^{-x}~dx=\lim_{a\rightarrow \infty} \frac{-1}{e^a}+1=0+1=1} . |
(b)
| Step 1: |
|---|
| First, we write Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_1^4 \frac{dx}{\sqrt{4-x}}=\lim_{a\rightarrow 4} \int_1^a\frac{dx}{\sqrt{4-x}}} . |
| Now, we proceed by Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u} -substitution. We let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u=4-x} . Then, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle du=-dx} . |
| Since the integral is a definite integral, we need to change the bounds of integration. |
| Plugging in our values into the equation Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u=4-x} , we get Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u_1=4-1=3} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u_2=4-a} . |
| Thus, the integral becomes |
| Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_1^4 \frac{dx}{\sqrt{4-x}}=\lim_{a\rightarrow 4} \int_3^{4-a}\frac{-du}{\sqrt{u}}} . |
| Step 2: |
|---|
| We integrate to get |
| Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_1^4 \frac{dx}{\sqrt{4-x}}=\lim_{a\rightarrow 4} \left.-2u^{\frac{1}{2}}\right|_{3}^{4-a}=\lim_{a\rightarrow 4}-2\sqrt{4-a}+2\sqrt{3}=2\sqrt{3}} |
| Final Answer: |
|---|
| (a) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1} |
| (b) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2\sqrt{3}} |