CAN TWO DIFFERENT SIZE SQUARES HAVE THE SAME PERIMETER ? EXPLAIN.
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Finite approximation method of estimating the area under the curve of the
given function makes use of rectangular approximation of the area.
The correct responses are;
i) The estimated area using a lower sum with two rectangles of equal width is <u>1,715 square units</u>.
ii) The estimated area using a lower sum with four rectangles of equal width is <u>3,001.25 square units</u>.
iii) The estimated area using an upper sum with two rectangles of equal width is<u> 8,575 square units</u>.
iv) The estimated area using a upper sum with four rectangles of equal width is <u>6,431.25 square units</u>.
Reasons:
The given function is f(x) = 5·x²
The given domain is x₀ to x₁₄
i) Estimate using lower sum with two rectangles of equal width;
f(0) = 0
f(7) = 5 × 7² = 245
A = 0 × 7 + 245 × 7 = 1,715
The estimated area using a lower sum with two rectangles of equal width
is <u>1,715 square units</u>.
ii) Estimate using lower sum with four rectangles of equal width;
f(0) = 0
f(3.5) = 5 × 3.5² = 61.25
f(7) = 5 × 7² = 245
f(10.5) = 5 × 10.5² = 551.25
A = 0 × 3.5 + 61.25 × 3.5 + 245 × 3.5 + 551.25 × 3.5 = 3,001.25
The estimated area using a lower sum with four rectangles of equal width is <u>3,001.25 square units</u>.
iii) Estimate using an upper sum with two rectangles of equal width;
f(7) = 5 × 7² = 245
f(14) = 5 × 14² = 980
A = 245 × 7 + 980 × 7 = 8575
The estimated area using an upper sum with two rectangles of equal width
is <u>8,575 square units</u>.
iv) Estimate using an upper sum with four rectangles of equal width;
f(3.5) = 5 × 3.5² = 61.25
f(7) = 5 × 7² = 245
f(10.5) = 5 × 10.5² = 551.25
f(14) = 5 × 14² = 980
A = 61.25 × 3.5 + 245 × 3.5 + 551.25 × 3.5 + 980 × 3.5 = 6,431.25
The estimated area using a upper sum with four rectangles of equal width
is <u>6,431.25 square units</u>.
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