Answer:
Area under the curve
f(x)=x², from x=1 to x=5 will be equal to
[tex]=\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}[/tex]
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
[tex]=\frac{1}{2} \times \text{base} \times \text{Height}\\\\A_{1},A_{2},A_{3},A_{4},\text{are four right triangles}\\\\A_{1}=\frac{1}{2} \times1 \times 3\\\\=\frac{3}{2}\\\\A_{2}=\frac{1}{2} \times 1 \times 5\\\\=\frac{5}{2}\\\\A_{3}=\frac{1}{2} \times 1 \times 7\\\\=\frac{7}{2}\\\\A_{4}=\frac{1}{2} \times 1 \times 9\\\\=\frac{9}{2}\\\\A_{1}+A_{2}+A_{3}+A_{4}=\frac{3}{2}+\frac{5}{2}+\frac{7}{2}+\frac{9}{2}\\\\=12[/tex]
Required Area of the Region
= Area of four Rectangles -Area of Triangle
=54 Square unit - 12 Square unit
=42 Square Unit
