Area between two curves integral

dBanji

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Hi everyone, please I need help with this problem. I am stuck. See below
Can someone please help with this problem: Approximate the points at which the graphs of f(x) = 4x^2 − 1 and g(x) = (1 + 4x^2)^(−3/2) intersect, and approximate the area (in units^2) between their graphs accurate to three decimal places.

I used a graphing calculator to plot the graph and found the point of intersections to be approx x=0.6 and -0.6. Then I found the antiderivative of f(x) as (4x^3)/3 -x. the antiderivative of g(x) I found to be [-(1+4x^2)^(-1/2)]/4x. I did not include +C because it's definite integral from -0.6 to 0.6. I subtracted the lower function f(x) from the upper function g(x) i.e. g(x) - f(x) and got 0.065 but that was wrong. What else should I do? I need someone to help, please.

Thanks.
 
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The approx intersections are at [MATH]x \approx \pm 0.56763[/MATH]. Next step is [MATH]\int_a^b (\textcolor{red}{top} - \textcolor{green}{bottom} ) dx = area[/MATH]. In your case the top function is [MATH]\textcolor{red}{g(x) = (1+4x^2)^{-\frac{3}{2}}}[/MATH] and the bottom function is [MATH]\textcolor{green}{f(x) = 4x^2-1}[/MATH]. The integral is as follows [MATH]\int_{-0.56763}^{0.56763} \left(g(x) -f(x)\right) \, dx = \int_{-0.56763}^{0.56763} \left((1+4x^2)^{-\frac{3}{2}} -(4x^2-1)\right) \, dx \approx 1.39794[/MATH]
IntGraphs.png

Short side question: Is there additionally asked for a method to determine the intersection point? I mean you can do that graphically, via calculator, but analytically or approximatively goes also...
 
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The approx intersections are at [MATH]x \approx \pm 0.56763[/MATH]. Next step is [MATH]\int_a^b (\textcolor{red}{top} - \textcolor{green}{bottom} ) dx = area[/MATH]. In your case the top function is [MATH]\textcolor{red}{g(x) = (1+4x^2)^{-\frac{3}{2}}}[/MATH] and the bottom function is [MATH]\textcolor{green}{f(x) = 4x^2-1}[/MATH]. The integral is as follows [MATH]\int_{-0.56763}^{0.56763} \left(g(x) -f(x)\right) \, dx = \int_{-0.56763}^{0.56763} \left((1+4x^2)^{-\frac{3}{2}} -(4x^2-1)\right) \, dx \approx 1.39794[/MATH]
View attachment 26621

Short side question: Is there additionally asked for a method to determine the intersection point? I mean you can do that graphically, via calculator, but analytically or approximatively goes also...
I cannot thank you enough. I am so grateful. The answer is correct now. I did not use the calculator to integrate rather I worked it out. I had the correct result for the f(x) but the g(x) was wrong. I think the problem was the integration process. How do you integrate g(x) without using a calculator.
Thank you for finding time to help. I appreciate it.

And what software did you use to type those integrations and graph? I'd like to have it as well for clarity of explanation or question.
 
No problem :) For the plot I used a software called Desmos, here is a link for the online plotter: https://www.desmos.com/calculator?lang=en

It is possible to integrate this via hand, but therefore one have to use for example the (not obvious) substitution [MATH]x = \frac{\tan(u)}{2}[/MATH]
 
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