Set up and then use limits and the formula:Σnᵢ=1 i2 =1/6 n(n + 1)(2n + 1)  to find the exact value of ʃ²0 3x2dx.

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lfryerda | High School Teacher | (Level 2) Educator

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Using the sum `sum_{i=1}^n i^2={n(n+1)(2n+1)}/6` to find the value of the integral `3int_0^2x^2dx` we need to change the integral into a sum of rectangles, and then let the width of each rectangle go to zero.

Consider the function `f(x)=3x^2`.  If we look at a rectangle divided into n equal intervals, then the width of each rectangle is `Delta x=2/n`. Then the `i^{th}` interval has initial position `x=i Deltax={2i}/n`, the final position is `x=(i+1) Delta x` and the rectangle height at the initial position is `3({2i}/n)^2={12i^2}/n^2`.  This means that the area of the rectangle is `Delta x cdot {12i^2}/n^2={24i^2}/n^3`.

Now the approximate area under the curve is the sum of all the rectangles.  This area only becomes exact once we take the limit as the width of each rectangle goes to zero, which is also the same as taking the limit as n goes to infinity.

The area of the rectangles is

`sum_{i=1}^n {24i^2}/n^3`    take the constants out of the sum

`={24}/n^3 sum_{n=1}^n i^2`   now use the sum provided

`={24}/n^3 cdot {n(n+1)(2n+1)}/6`   now factor an n out of the RHS

`={4}/n^3 cdot n^3(1+1/n)(2+1/n)`

`=4(1+1/n)(2+1/n)`

Now we can take the limit as which gives the exact area

`lim_{n->infty} 4(1+1/n)(2+1/n)`

`=4(1+0)(2+0)`

`=8`

So the exact area under the curve is 8 square units.

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cspanutius | Student, Undergraduate | (Level 1) Salutatorian

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the question should read: Set up and then use limits and the formula:Σ(superscript n)(subscript i=1) i^2 =1/6 n(n + 1)(2n + 1)  to find the exact value of ʃ^(superscript ²)(subscript 0) 3x^2dx

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