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Anonymous' question at Yahoo! Answers regarding the area bounded by two curves
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Hello Anonymous,
We are given the two curves:
\(\displaystyle f(x)=x^3+4x\)
\(\displaystyle g(x)=6x^2-x\)
And we are asked to determine the area of the region enlocsed by the graphs of $f$ and $g$.
The first thing we want to do is find the $x$-coordinates of their intersections. To do this, we may equate the two functions and solve for $x$:
\(\displaystyle x^3+4x=6x^2-x\)
Arrange as a polynomial in standard form:
\(\displaystyle x^3-6x^2+5x=0\)
Factor:
\(\displaystyle x(x-1)(x-5)=0\)
Thus, by the zero-factor property, we find:
\(\displaystyle x=0,\,1,\,5\)
Next, we want to see which function is greater within the two intervals created.
i) On the interval \(\displaystyle [0,1]\):
Let's use the test value \(\displaystyle x=\frac{1}{2}\)
\(\displaystyle f\left(\frac{1}{2} \right)=\left(\frac{1}{2} \right)^3+4\left(\frac{1}{2} \right)=\frac{1}{8}+2=\frac{17}{8}\)
\(\displaystyle g\left(\frac{1}{2} \right)=6\left(\frac{1}{2} \right)^2-\left(\frac{1}{2} \right)=\frac{3}{2}-\frac{1}{2}=1\)
We may conclude then that on this interval:
\(\displaystyle f(x)\ge g(x)\)
ii) On the interval \(\displaystyle [1,5]\):
Let's use the test value \(\displaystyle x=3\)
\(\displaystyle f(3)=(3)^3+4(3)=27+12=39\)
\(\displaystyle g(3)=6(3)^2-(3)=54-3=51\)
We may conclude then that on this interval:
\(\displaystyle g(x)\ge f(x)\)
Hence, the area $A$ bounded by the two curves is given by:
\(\displaystyle A=\int_0^1 f(x)-g(x)\,dx+\int_1^5 g(x)-f(x)\,dx\)
\(\displaystyle A=\int_0^1 x^3-6x^2+5x\,dx-\int_1^5 x^3-6x^2+5x\,dx\)
Applying the FTOC, we obtain:
\(\displaystyle A=\left[\frac{1}{4}x^4-2x^3+\frac{5}{2}x^2 \right]_0^1-\left[\frac{1}{4}x^4-2x^3+\frac{5}{2}x^2 \right]_1^5\)
\(\displaystyle A=\left(\frac{1}{4}-2+\frac{5}{2} \right)-\left(\left(\frac{1}{4}(5)^4-2(5)^3+\frac{5}{2}(5)^2 \right)-\left(\frac{1}{4}-2+\frac{5}{2} \right) \right)\)
\(\displaystyle A=2\left(\frac{1}{4}-2+\frac{5}{2} \right)-5^2\left(\frac{1}{4}(5)^2-2(5)+\frac{5}{2} \right)\)
\(\displaystyle A=2\cdot\frac{3}{4}-25\cdot\left(-\frac{5}{4} \right)=\frac{3}{2}+\frac{125}{4}=\frac{131}{4}\)
We are given the two curves:
\(\displaystyle f(x)=x^3+4x\)
\(\displaystyle g(x)=6x^2-x\)
And we are asked to determine the area of the region enlocsed by the graphs of $f$ and $g$.
The first thing we want to do is find the $x$-coordinates of their intersections. To do this, we may equate the two functions and solve for $x$:
\(\displaystyle x^3+4x=6x^2-x\)
Arrange as a polynomial in standard form:
\(\displaystyle x^3-6x^2+5x=0\)
Factor:
\(\displaystyle x(x-1)(x-5)=0\)
Thus, by the zero-factor property, we find:
\(\displaystyle x=0,\,1,\,5\)
Next, we want to see which function is greater within the two intervals created.
i) On the interval \(\displaystyle [0,1]\):
Let's use the test value \(\displaystyle x=\frac{1}{2}\)
\(\displaystyle f\left(\frac{1}{2} \right)=\left(\frac{1}{2} \right)^3+4\left(\frac{1}{2} \right)=\frac{1}{8}+2=\frac{17}{8}\)
\(\displaystyle g\left(\frac{1}{2} \right)=6\left(\frac{1}{2} \right)^2-\left(\frac{1}{2} \right)=\frac{3}{2}-\frac{1}{2}=1\)
We may conclude then that on this interval:
\(\displaystyle f(x)\ge g(x)\)
ii) On the interval \(\displaystyle [1,5]\):
Let's use the test value \(\displaystyle x=3\)
\(\displaystyle f(3)=(3)^3+4(3)=27+12=39\)
\(\displaystyle g(3)=6(3)^2-(3)=54-3=51\)
We may conclude then that on this interval:
\(\displaystyle g(x)\ge f(x)\)
Hence, the area $A$ bounded by the two curves is given by:
\(\displaystyle A=\int_0^1 f(x)-g(x)\,dx+\int_1^5 g(x)-f(x)\,dx\)
\(\displaystyle A=\int_0^1 x^3-6x^2+5x\,dx-\int_1^5 x^3-6x^2+5x\,dx\)
Applying the FTOC, we obtain:
\(\displaystyle A=\left[\frac{1}{4}x^4-2x^3+\frac{5}{2}x^2 \right]_0^1-\left[\frac{1}{4}x^4-2x^3+\frac{5}{2}x^2 \right]_1^5\)
\(\displaystyle A=\left(\frac{1}{4}-2+\frac{5}{2} \right)-\left(\left(\frac{1}{4}(5)^4-2(5)^3+\frac{5}{2}(5)^2 \right)-\left(\frac{1}{4}-2+\frac{5}{2} \right) \right)\)
\(\displaystyle A=2\left(\frac{1}{4}-2+\frac{5}{2} \right)-5^2\left(\frac{1}{4}(5)^2-2(5)+\frac{5}{2} \right)\)
\(\displaystyle A=2\cdot\frac{3}{4}-25\cdot\left(-\frac{5}{4} \right)=\frac{3}{2}+\frac{125}{4}=\frac{131}{4}\)