what is the area of the region ?

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can we do this without integration ( breaking the area into triangles & rectangles etc ) ?

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Note the three similar right triangles. One below the top line, another below the bottom line and one above the $y=1$

The area of the biggest right triangle is $\frac{\frac{3}2\frac32}2=\frac98$
You have to subtract the area of other two (white ones) from this one,
The bottom one has $\frac{\frac{3}4\frac34}2=\frac9{32}$
The top one has $\frac{\frac{1}2\frac12}2=\frac1{8}$

Finally you have $\frac98-\frac{1}{8}-\frac9{32}=\frac{23}{32}$