D decoy808 Feb 2009 101 2 May 19, 2010 #1 can anyone help me with the process to the question. points i need to research rather than the answers. many thanks The function \(\displaystyle y=2(x-1)(x-4)^2\) (a) find the values of A and B (b) what is the size of the shaded part

can anyone help me with the process to the question. points i need to research rather than the answers. many thanks The function \(\displaystyle y=2(x-1)(x-4)^2\) (a) find the values of A and B (b) what is the size of the shaded part

skeeter MHF Helper Jun 2008 16,216 6,764 North Texas May 19, 2010 #2 decoy808 said: can anyone help me with the process to the question. points i need to research rather than the answers. many thanks The function \(\displaystyle y=2(x-1)(x-4)^2\) (a) find the values of A and B (b) what is the size of the shaded part Click to expand... should be clear that A = 1 and B = 4 (roots of the cubic, one w/ multiplicity two) area = \(\displaystyle \int_1^4 2(x-1)(x-4)^2 \, dx\) expand the cubic, then integrate and evaluate the definite integral using the fundamental theorem of calculus.

decoy808 said: can anyone help me with the process to the question. points i need to research rather than the answers. many thanks The function \(\displaystyle y=2(x-1)(x-4)^2\) (a) find the values of A and B (b) what is the size of the shaded part Click to expand... should be clear that A = 1 and B = 4 (roots of the cubic, one w/ multiplicity two) area = \(\displaystyle \int_1^4 2(x-1)(x-4)^2 \, dx\) expand the cubic, then integrate and evaluate the definite integral using the fundamental theorem of calculus.

D DrDank May 2010 20 8 May 19, 2010 #3 \(\displaystyle \) A and B occur when y=0 \(\displaystyle y=2(x-1)(x-4)^2\) \(\displaystyle 0=2(x-1)(x-4)^2\) \(\displaystyle 0=(x-1) \) and \(\displaystyle 0=(x-4)\) Therefore... \(\displaystyle A=1\) and \(\displaystyle B=4\) The Area is.. \(\displaystyle \int2(x-1)(x-4)^2dx\) \(\displaystyle 2\int(x-1)(x^2-8x+16)dx\) \(\displaystyle 2\int(x^3-7x^2+24x-16)dx\) Limits of integration are 1 to 4

\(\displaystyle \) A and B occur when y=0 \(\displaystyle y=2(x-1)(x-4)^2\) \(\displaystyle 0=2(x-1)(x-4)^2\) \(\displaystyle 0=(x-1) \) and \(\displaystyle 0=(x-4)\) Therefore... \(\displaystyle A=1\) and \(\displaystyle B=4\) The Area is.. \(\displaystyle \int2(x-1)(x-4)^2dx\) \(\displaystyle 2\int(x-1)(x^2-8x+16)dx\) \(\displaystyle 2\int(x^3-7x^2+24x-16)dx\) Limits of integration are 1 to 4