Area bounded by curve ,x & y axes using integration.

Here is the question: The graph y= 2x^3-15x^2+36x-28 has a double root at x=2. Find the area bounded by the graph, the x axis, the y axis and the line x=2. The answer is 16 ( modulus). My question is if you integrate the cubic between x=2 and x=0, you get a finite answer ( 16) , but when you see the graph ( curving up through y=-28 to reach a max turning point at x=2 , then falling ( but this bit doesn't matter)). What matters is that for me UNDER.the cubic ( between 2 & 0 )the area just stretches down forever. Hence i don't understand how the specific area can be calculated without creating another boundary of y=-28. Much appreciated if anyone can help me here. thanks

Reply 1

mqb2766

Original post

by Yerffoeg

The area doesnt stretch on forever (the y axis or x=0 is not a vertical asymptote). The y intercept is -28 (at x=0), so youre integrating between that and (2,0). The x and y axes specify the region along with the curve and up to x=2, as the question says.

Edit -if youre thinking about integration as area under a graph, thats wrong. Its the area between the curve and the x-axis. When the curve is positive, its the same thing, but when the curve is negative the integral reprsents the (negative) area above the curve and under the x-axis.