Answer:
Area under the curve
f(x)=x², from x=1 to x=5 will be equal to
![=\int\limits^5_1 {x^2} \, dx \\\\=[\frac{x^3}{3}]\left \{ {{x=5} \atop {x=1}} \right.\\\\=\frac{5^3}{3}-\frac{1^3}{3}\\\\ =\frac{125}{3}-\frac{1}{3}\\\\=\frac{124}{3}](https://tex.z-dn.net/?f=%3D%5Cint%5Climits%5E5_1%20%7Bx%5E2%7D%20%5C%2C%20dx%20%5C%5C%5C%5C%3D%5B%5Cfrac%7Bx%5E3%7D%7B3%7D%5D%5Cleft%20%5C%7B%20%7B%7Bx%3D5%7D%20%5Catop%20%7Bx%3D1%7D%7D%20%5Cright.%5C%5C%5C%5C%3D%5Cfrac%7B5%5E3%7D%7B3%7D-%5Cfrac%7B1%5E3%7D%7B3%7D%5C%5C%5C%5C%20%3D%5Cfrac%7B125%7D%7B3%7D-%5Cfrac%7B1%7D%7B3%7D%5C%5C%5C%5C%3D%5Cfrac%7B124%7D%7B3%7D)
Area of the curve to the nearest integer=41 square units
⇒Using Rectangles Only
Area of Rectangle=Length × Breadth
→x=1, gives, y=1
→x=2, gives, y=4
Area of First rectangle =1×4=4 Square unit
→x=3, gives, y=9
Area of Second rectangle =1×9=9 Square unit
→x=4, gives, y=16
→x=5, gives, y=25
Area of Third rectangle =1×16=16 Square unit
Area of fourth rectangle =1×25=25 Square unit
Area of Rectangles=4+9+16+25
=54 square units
Area of Triangle
Required Area of the Region
= Area of four Rectangles -Area of Triangle
=54 Square unit - 12 Square unit
=42 Square Unit