Question

Find the area of the region bound on the left by the line x = 1, on the right by x = 2, the x-axis, and the curve.

Answer 1

\[\frac{ x-3 }{ x^3+x^2}\]

Answer 2

is the equation of the curve

Answer 3

\[\int\limits_{1}^{2}(-\frac{ x-3 }{ x^3+x^2 })dx\]

Answer 4

Is this right?

Answer 5

Where did the negative sign come from?

Answer 6

0-the equation because the x axis is on top

Answer 7

Oh, right! This integral requires partial fraction decomposition.
\[-\frac{x-3}{x^3+x^2}=-\frac{x-3}{x^2(x+1)}=-\frac{A}{x}-\frac{B}{x^2}-\frac{C}{x+1}\]
\[x-3=Ax(x+1)+B(x+1)+Cx^2\\
x-3=(A+C)x^2+(A+B)x+B\]
Matching up coefficients, you have
\[\begin{cases}A+C=0\\A+B=1\\B=-3\end{cases}\]
Solve for the constants, plug them into the above fractions, then integrate term by term.

Answer 8

so B=-3, A=4, C=-4

Answer 9

\[-(4lnx-3lnx^2-4\ln(x+1))\]

Answer 10

right?

Answer 11

2nd term should be -3x^-1

Answer 12

oh right! thank you

Answer 13

i'm not sure about the -ve signs for each term. doesnt look right

Answer 14

I got the right answer

Answer 15

ok good. so the signs are correct, well done