How do you evaluate the definite integral by the limit definition given #int 4dx# from [-a,a]?

Question

Paisley Johnson · Accepted Answer

# int \ 4 \ dx = 8a #

We seek:

# A = int_(-a)^(a) \ 4 \ dx #

We assume wlog that a #gt 0#

Let us use our common sense, we know that a definite:

# A = int_alpha^beta \ f(x) \ dx #

represents the area under the curve #y=f(x)# between #x=alpha# and #x=beta#, thus the integral we seek represents the area under the fixed curve #y=4# between #x=-a# and #x=a#, which is a rectangle of height #4# and width #2a#.

Hence,

# int \ 4 \ dx = 8a #

We can establish the same result by using the definition of the integral in term of a riemann sum. We can estimate this area under the curve using thin rectangles. The more rectangles we use, the better the approximation gets, and calculus deals with the infinite limit of a finite series of infinitesimally thin rectangles.

That is

# int_a^b \ f(x) \ dx = lim_(n rarr oo) (b-a)/n sum_(i=1)^n \ f(a + i(b-a)/n)#

And we partition the interval #[a,b]# equally spaced using:

# Delta = {a+0((b-a)/n), a+1((b-a)/n), ..., a+n((b-a)/n) } # # \ \ \ = {a, a+1((b-a)/n),a+2((b-a)/n), ... ,b } #

Here we have #f(x)=4# and so we partition the interval #[-a,a]# using:

# Delta = {-a, -a+2(2a)/n, -a+2 (2a)/n, -a+3 (2a)/n, ..., a } #

And so:

# I = int_(-a)^(a) \ 4 \ dx # # \ \ = lim_(n rarr oo) (2a)/n sum_(i=1)^n \ f(-a+i*(2a)/n)# # \ \ = lim_(n rarr oo) (2a)/n sum_(i=1)^n \ 4# # \ \ = lim_(n rarr oo) (2a)/n (4n)# # \ \ = lim_(n rarr oo) 8a# # \ \ = 8a#

Using Calculus

If we use Calculus and our knowledge of integration to establish the answer, for comparison, we get:

# int_(-a)^(a) \ 4 \ dx = [ 4x ]_(-a)^(a) #

# " " = 4[ x ]_(-a)^(a) # # " " = 4{ (a) - (-a) } # # " " = 4(2a) # # " " = 8a #

Samantha Adkison · Answer

undefined To evaluate the definite integral ∫(−a to a) 4dx using the limit definition, you follow these steps:

1. Write down the limit definition of the definite integral:

∫(−a to a) f(x)dx = lim(n→∞) [Σ(i=1 to n) f(x_i*) Δx]

2. For this integral, f(x) = 4.

3. Determine the interval size Δx. Since the interval is from −a to a, the width of each subinterval is 2a/n.

4. Choose sample points x_i* in each subinterval. Typically, you can choose the midpoint of each subinterval.

5. Plug the values of f(x) and Δx into the summation:

lim(n→∞) [Σ(i=1 to n) 4 Δx]

6. Simplify the expression inside the summation to get:

lim(n→∞) [4 * 2a/n * n]

7. Cancel out n from the expression to obtain:

lim(n→∞) [8a]

8. Evaluate the limit:

lim(n→∞) [8a] = 8a

9. So, the value of the definite integral ∫(−a to a) 4dx is 8a.