area under the curve

**ahawk1** · February 3rd 2009, 08:26 PM

Find the area of the region bounded by the given curves.

i got the answer

is this correct? if not can u show me how u did it?

**Soroban** · February 3rd 2009, 09:30 PM

Hello, ahawk1!

I wish you'd shown us your work . . .

Find the area of the region bounded by the given curves: . $\begin{array}{cccc}y &=& 35\ln x & {\color{blue}[1]}\\ y&=&x\ln x & {\color{blue}[2]}\end{array}$

Find the intersections of the curves . . .

$x\ln x \:=\:35\ln x \quad\Rightarrow\quad x\ln x - 35\ln x \:=\:0 \quad\Rightarrow\quad \ln x\bigg(x - 35\bigg) \:=\:0$

And we have: . $\begin{array}{ccccccc}\ln x \:=\:0 & \Rightarrow & x \:=\:1 \\ x-35 \:=\:0 & \Rightarrow & x \:=\:35 \end{array}$

On the interval [1, 35], [1] is the upper curve.

The integral is: . $\int^{35}_1\bigg[35\ln x - x\ln x\bigg]\,dx \;=\;35\int^{35}_1\ln x\,dx - \int^{35}_1 x\ln x\,dx<br /> <br />$

Both integrals require integration-by-parts.

We get: . $35x(\ln x - 1) - \tfrac{1}{4}x^2(2\ln x - 1)\,\bigg]^{35}_1$ . . . . etc.

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