# How do you evaluate the definite integral by the limit definition given #int (a-absx)dx# from [-a,a]?

The answer using limit definition has already appeared. I am giving

other methods.

In the left half, it is a-(-x)=a+x and the integral is

In the second half, the integrand is a-x and the integral is

See the graph for a = 2.

graph{(2-|x|-y)y((x-2)^2+y^2-.05)((x+2)^2+y^2-.05)(x^2+(y-2)^2-.05)=0 [-10, 10, -5, 5]}

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See below

Because of the absolute value it might be sensible to split the integration as follows:

For the first part, we are looking for a summation in form

Repeat for the other interval or use symmetry.

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To evaluate the definite integral ( \int_{-a}^{a} (a - |x|) , dx ) using the limit definition, we first express it as a limit of Riemann sums:

[ \int_{-a}^{a} (a - |x|) , dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i) \cdot \Delta x ]

where ( f(x) = a - |x| ), ( \Delta x = \frac{b - a}{n} ), and ( x_i ) are sample points within each subinterval.

Next, we partition the interval ([-a, a]) into (n) subintervals of equal width:

[ \Delta x = \frac{2a}{n} ]

and choose sample points ( x_i ) in each subinterval. Since the function ( f(x) = a - |x| ) is symmetric about the origin, we can simplify the calculation by considering only one half of the interval, for example, ([0, a]).

Then, ( x_i ) for the (i)-th subinterval is ( x_i = \frac{2ai}{n} ).

Now, we evaluate ( f(x_i) ) at each sample point:

[ f(x_i) = a - \left| \frac{2ai}{n} \right| ]

[ = a - \frac{2ai}{n} ]

Finally, we substitute these values into the Riemann sum formula and take the limit as ( n ) approaches infinity:

[ \lim_{n \to \infty} \sum_{i=1}^{n} \left( a - \frac{2ai}{n} \right) \frac{2a}{n} ]

[ = \lim_{n \to \infty} \frac{4a^2}{n^2} \sum_{i=1}^{n} \left( n - 2i \right) ]

[ = \lim_{n \to \infty} \frac{4a^2}{n^2} \left( n^2 - 2n \frac{n+1}{2} \right) ]

[ = \lim_{n \to \infty} \frac{4a^2}{n^2} \left( n^2 - n^2 - n \right) ]

[ = \lim_{n \to \infty} \frac{4a^2}{n^2} \left( -n \right) ]

[ = \lim_{n \to \infty} \left( -\frac{4a^2}{n} \right) ]

[ = 0 ]

Therefore, the definite integral ( \int_{-a}^{a} (a - |x|) , dx ) evaluates to ( 0 ).

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