Discover the answers you need at Westonci.ca, a dynamic Q&A platform where knowledge is shared freely by a community of experts. Get the answers you need quickly and accurately from a dedicated community of experts on our Q&A platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
Sagot :
Find where the two curves meet.
[tex]x^3 - x^2 + x + 8 = 5x^2 - 7x + 8 \\\\ \implies x^3 - 6x^2 + 8x = 0 \\\\ \implies x (x-2) (x - 4)= 0 \implies x=0, x=2, x=4[/tex]
The area between the curves is
[tex]\displaystyle \int_0^4 \left|\left(x^3-x^2+x+8\right) - \left(5x^2 - 7x + 8\right)\right| \, dx = \int_0^4 \left|x(x-2)(x-4)\right| \, dx[/tex]
When [tex]x[/tex] is between 0 and 2, [tex]x(x-2)(x-4)[/tex] is positive; when [tex]x[/tex] is between 2 and 4, [tex]x(x-2)(x-4)[/tex] is negative. So we split the integral at [tex]x=2[/tex] to get
[tex]\displaystyle \int_0^2 x(x-2)(x-4) \, dx - \int_2^4 x(x-2)(x-4)\,dx[/tex]
In the second integral, substitute [tex]y=x-2[/tex] to get
[tex]\displaystyle \int_0^2 x(x-2)(x-4) \, dx - \int_0^2 (y+2)y(y-2)\,dy[/tex]
[tex]\displaystyle \int_0^2 x(x-2) \bigg((x-4) - (x+2)\bigg) \, dx[/tex]
[tex]\displaystyle -6 \int_0^2 x(x-2) \, dx[/tex]
[tex]\displaystyle 6 \int_0^2 \left(2x - x^2\right) \, dx[/tex]
[tex]\displaystyle 6 \left(x^2 - \frac13x^3\right)\bigg|_0^2 = \boxed{8}[/tex]
Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.