Hello Anonymous,
We are given the two curves:
$$f(x)=x^3+4x$$
$$g(x)=6x^2-x$$
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 $x$:
$$x^3+4x=6x^2-x$$
Arrange as a polynomial in standard form:
$$x^3-6x^2+5x=0$$
Factor:
$$x(x-1)(x-5)=0$$
Thus, by the zero-factor property, we find:
$$x=0,\,1,\,5$$
Next, we want to see which function is greater within the two intervals created.
i) On the interval $$[0,1]$$:
Let's use the test value $$x=\frac{1}{2}$$
$$f\left(\frac{1}{2} \right)=\left(\frac{1}{2} \right)^3+4\left(\frac{1}{2} \right)=\frac{1}{8}+2=\frac{17}{8}$$
$$g\left(\frac{1}{2} \right)=6\left(\frac{1}{2} \right)^2-\left(\frac{1}{2} \right)=\frac{3}{2}-\frac{1}{2}=1$$
We may conclude then that on this interval:
$$f(x)\ge g(x)$$
ii) On the interval $$[1,5]$$:
Let's use the test value $$x=3$$
$$f(3)=(3)^3+4(3)=27+12=39$$
$$g(3)=6(3)^2-(3)=54-3=51$$
We may conclude then that on this interval:
$$g(x)\ge f(x)$$
Hence, the area $A$ bounded by the two curves is given by:
$$A=\int_0^1 f(x)-g(x)\,dx+\int_1^5 g(x)-f(x)\,dx$$
$$A=\int_0^1 x^3-6x^2+5x\,dx-\int_1^5 x^3-6x^2+5x\,dx$$
Applying the FTOC, we obtain:
$$A=\left[\frac{1}{4}x^4-2x^3+\frac{5}{2}x^2 \right]_0^1-\left[\frac{1}{4}x^4-2x^3+\frac{5}{2}x^2 \right]_1^5$$
$$A=\left(\frac{1}{4}-2+\frac{5}{2} \right)-\left(\left(\frac{1}{4}(5)^4-2(5)^3+\frac{5}{2}(5)^2 \right)-\left(\frac{1}{4}-2+\frac{5}{2} \right) \right)$$
$$A=2\left(\frac{1}{4}-2+\frac{5}{2} \right)-5^2\left(\frac{1}{4}(5)^2-2(5)+\frac{5}{2} \right)$$
$$A=2\cdot\frac{3}{4}-25\cdot\left(-\frac{5}{4} \right)=\frac{3}{2}+\frac{125}{4}=\frac{131}{4}$$
