Question

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

Answer 1

See below

Explanation:

`int_(-a)^a a-absx dx`

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

`= ( int_(-a)^0 + int_(0)^a) a-absx \ \ dx`

`= int_(-a)^0 a + x \ dx + int_(0)^a a-x \ \ dx`

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

`= int_(-a)^0 a + x \ dx = lim_(n to infty) sum_(i = 1)^(n) f(x_i) Delta x`

We split the interval into `n` rectangles of width `Delta x` such that:

`Delta x = (0 - (- a))/n = a/n`

The right side of the ith rectangle is located at
`x_i = - a + i Delta x`

`f(x_i) = a + (- a + i Delta x) = i Delta x`

`implies lim_(n to infty) sum_(i = 1)^(n) i Delta x * Delta x`

`= lim_(n to infty) sum_(i = 1)^(n) (i a^2)/n^2`

`= color(red)(a^2) lim_(n to infty) color(blue)(1/n^2) color(cyan)( sum_(i = 1)^(n) i )`

`= a^2 lim_(n to infty) 1/n^2 * color(cyan)( (n^2+n)/2)`

`= a^2 lim_(n to infty) 1/2 + 1/(2n)`

`= a^2 /2`

This is what we expect as we are evaluating the area of a triangle of height and base `a`.

Repeat for the other interval or use symmetry.

Answer 2

`a^2`. The integral is area under the curve and the graph forms an isosceles `triangle`, with base 2a on the x-axis. The area of the `triangle` is `1/2(2a)(a)=a^2`..

Explanation:

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

other methods.

The integrand is continuous in `[-a, a]`

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

`[ax+x^2/2],` between `x = -a and 0`, and this is

`[0]-[-a^2+a^2/2]=a^2/2`.

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

`[ax-x^2/2]`, between the limits x = 0 and a, and this part is

`[a^2-a^2/2]-[0]=a^2/2`.

Adding, the integral is `a^2`.

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]}