Science:Math Exam Resources/Courses/MATH101/April 2018/Question 03 (i)

MATH101 April 2018

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Question 03 (i)
Determine explicitly the area of the bounded region between the curves ${\textstyle y=x^{4}}$ and ${\textstyle y=4-3x^{2}}$ .

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
The area of the bounded region between the curves ${\textstyle y=f(x)}$ and ${\textstyle y=g(x)}$ , where two curves intersect at $x=a$ and $x=b$ and $f(x)\geq g(x)$ on the interval $(a,b)$ , is $\int _{a}^{b}(f(x)-g(x))\ dx.$

Hint 2
Observe that both ${\textstyle y=x^{4}}$ and ${\textstyle y=4-3x^{2}}$ are even functions.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work.

If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work.
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Solution
Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. For the convenience, let $f(x)=4-3x^{2}$ and $g(x)=x^{4}$ . First, we find the intersection points of the given curves. It is enough to solve $f(x)=g(x)\iff x^{4}+3x^{2}-4=0\iff (x^{2}+4)(x^{2}-1)=0\iff x^{2}-1=0\iff x=\pm 1.$ The second equivalence can be obtained by putting $u=x^{2}$ and solving $u^{2}+3u-4=0$ . The third equivalence holds because $x^{2}+4$ is nonzero for all x (indeed, $x^{2}+4\geq 4$ ). Observe that $f(x)\geq g(x)$ on $(-1,1)$ . (We can check this easily by comparing the values of two functions at some point in the interval, like $4=f(0)\geq g(0)=0$ . This is because both functions are continuous and there's no intersection points between $-1$ and $1$ .) Also, both functions $f$ and $g$ are even. Therefore, by the Hint 1, we have the area $\int _{-1}^{1}(4-3x^{2}-x^{4})dx=2\int _{0}^{1}(4-3x^{2}-x^{4})dx=2\left[4x-x^{3}-{\frac {1}{5}}x^{5}\right]_{x=0}^{x=1}=2\left[\left(4-1-{\frac {1}{5}}\right)-\left(0-0-0\right)\right]={\frac {28}{5}}.$ Answer: $\color {blue}{\frac {28}{5}}$