How Does Integration Apply to Subtraction of Function Areas?

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Let [PLAIN][PLAIN]http://msr02.math.mcgill.ca/webwork2_files/jsMath/fonts/cmex10/alpha/144/char5A.png f(x)dx=5 a=7, b= 13[PLAIN]http://msr02.math.mcgill.ca/webwork2_files/jsMath/fonts/cmmi10/alpha/144/char3B.png [PLAIN][PLAIN]http://msr02.math.mcgill.ca/webwork2_files/jsMath/fonts/cmex10/alpha/144/char5A.png f(x)dx=3 a=7, b=9[PLAIN]http://msr02.math.mcgill.ca/webwork2_files/jsMath/fonts/cmmi10/alpha/144/char3B.png [PLAIN][PLAIN]http://msr02.math.mcgill.ca/webwork2_files/jsMath/fonts/cmex10/alpha/144/char5A.png f(x)dx=5 a=11,b=13

Find [PLAIN][PLAIN]http://msr02.math.mcgill.ca/webwork2_files/jsMath/fonts/cmex10/alpha/144/char5A.png f(x)dx[FONT=.LucidaGrandeUI]= a=9 b = 11 ==== I figureed out it will be 3+5-5=3 therefore it is =-3
and [PLAIN][PLAIN]http://msr02.math.mcgill.ca/webwork2_files/jsMath/fonts/cmex10/alpha/144/char5A.png (5f(x)−3)dx= a=11 b = 9 I am lost. I know it will be 3 but then its 5f(x)-3 which I don't get.a=bottom b=top boundaries
 
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$$\int_a^b 5f(x)-3\; dx = 5\int_a^b f(x)\; dx - 3\int_a^b\;dx$$
 
Homework-type problems should be posted in the homework & coursework sections, not in the technical math sections. I have moved this thread to the appropriate forum section.
 
mshiddensecret said:
Let [PLAIN][PLAIN]http://msr02.math.mcgill.ca/webwork2_files/jsMath/fonts/cmex10/alpha/144/char5A.png f(x)dx=5 a=7, b= 13[PLAIN]http://msr02.math.mcgill.ca/webwork2_files/jsMath/fonts/cmmi10/alpha/144/char3B.png [PLAIN][PLAIN]http://msr02.math.mcgill.ca/webwork2_files/jsMath/fonts/cmex10/alpha/144/char5A.png f(x)dx=3 a=7, b=9[PLAIN]http://msr02.math.mcgill.ca/webwork2_files/jsMath/fonts/cmmi10/alpha/144/char3B.png [PLAIN][PLAIN]http://msr02.math.mcgill.ca/webwork2_files/jsMath/fonts/cmex10/alpha/144/char5A.png f(x)dx=5 a=11,b=13

Find [PLAIN][PLAIN]http://msr02.math.mcgill.ca/webwork2_files/jsMath/fonts/cmex10/alpha/144/char5A.png f(x)dx= a=9 b = 11 ==== I figureed out it will be 3+5-5=3 therefore it is =-3
and [PLAIN][PLAIN]http://msr02.math.mcgill.ca/webwork2_files/jsMath/fonts/cmex10/alpha/144/char5A.png (5f(x)−3)dx= a=11 b = 9 I am lost. I know it will be 3 but then its 5f(x)-3 which I don't get.a=bottom b=top boundaries

Your problem statement and work are just about incomprehensible. My best guess is that this is the problem statement.
$$\int_7^{13}f(x)dx = 5 $$
$$\int_7^{9}f(x)dx = 3 $$
$$\int_{11}^{13}f(x)dx = 5 $$
To answer the stated questions, use the rule that says
$$\int_a^b f(x)dx + \int_b^c f(x)dx = \int_a^c f(x)dx$$
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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