MHB Area between Curves: Find 1st Quadrant

[karush]

Find the area of the region in the first quadrant
bounded by the line $y=x$, the line $x=2$, the curve $y=\frac{1}{x^2}$

$$\int_{1}^{2}\left(x-\frac{1}{{x}^{2}}\right) \,dx$$
just seeing if this is the way to go.

View attachment 4495

 

[Ackbach]

Looks good to me!

 

[karush]

The rest was easy the Area $ = 1$

 

[MarkFL]

Let's just see about that...

$$\int_1^2\left(x-\frac{1}{{x}^{2}}\right)\,dx=\left[\frac{x^2}{2}+\frac{1}{x}\right]_1^2=\left(2+\frac{1}{2}\right)-\left(\frac{1}{2}+1\right)=1$$

Yep...you are correct. (Mmm)