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

Question

Emilia Aguilar · Accepted Answer

#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#..

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]}

Scarlett Allison · Answer

See below

#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.

Ezekiel Alexander · Answer

undefined 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 $.