How do you use limits to evaluate #int (x+2)dx# from [1,4]?

Question

Emma Aldridge · Accepted Answer

# int_1^4 \ (x+2) \ dx = 13.5#

By definition of an integral, then # int_a^b \ f(x) \ dx # represents the area under the curve #y=f(x)# between #x=a# and #x=b#. 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)# Here we have #f(x)=x+2# and we partition the interval #[1,4]# using #Delta = {1, 1+3*1/n, 1+3*2/n, ..., 1+3*n/n }# And so: # I = int_1^4 \ (x+2) \ dx # # \ \ = lim_(n rarr oo) 3/n sum_(i=1)^n \ f(1+3*i/n)# # \ \ = lim_(n rarr oo) 3/n sum_(i=1)^n \ ((1+3*i/n)+2) # # \ \ = lim_(n rarr oo) 3/n sum_(i=1)^n \ (3+(3i)/n) # # \ \ = lim_(n rarr oo) 3/n {sum_(i=1)^n 3+ sum_(i=1)^n (3i)/n }# # \ \ = lim_(n rarr oo) 3/n {3n+ 3/nsum_(i=1)^n i }# Using the standard summation formula: # sum_(r=1)^n r = 1/2n(n+1) # we have: # I = lim_(n rarr oo) 3/n {3n+ 3/n*1/2n(n+1) }# # \ \ = lim_(n rarr oo) 3/n {3n+ 3/2(n+1) }# # \ \ = lim_(n rarr oo) 3/n*3/2 {2n+ (n+1) }# # \ \ = lim_(n rarr oo) 9/(2n) (3n+1)# # \ \ = lim_(n rarr oo) (27/2+9/(2n)) # # \ \ = 27/2# # \ \ = 13.5#

Emma Aldridge · Answer

Perform the integration as if the integral were indefinite but without a constant of integration.
Subtract the expression evaluated at the lower limit from the expression evaluated at the upper limit.

Given: #int_1^4 (x+2)dx# Perform the integration as if the integral were indefinite but without a constant of integration: #int_1^4 (x+2)dx= {:x^2/2+2x]_1^4# Subtract the resulting expression evaluated at the lower limit from the expression evaluated at the upper limit: #int_1^4 (x+2)dx= (4^2/2+2(4))- (1^2/2+2(1))# #int_1^4 (x+2)dx= 8+8- 1/2-2# #int_1^4 (x+2)dx= 13.5#

Samantha Adkison · Answer

undefined To evaluate the integral ∫(x+2)dx from 1 to 4 using limits, you can follow these steps:

1. Integrate the function (x+2) with respect to x to find the antiderivative.
2. Evaluate the antiderivative at the upper limit of integration (4) and subtract the value of the antiderivative at the lower limit of integration (1).
3. This gives the definite integral of the function over the interval [1, 4].

Let's go through the steps:

1. The antiderivative of (x+2) with respect to x is (1/2)x^2 + 2x + C, where C is the constant of integration.
2. Evaluate the antiderivative at the upper limit (4):
   (1/2)(4)^2 + 2(4) = 8 + 8 = 16
   Then, subtract the value of the antiderivative at the lower limit (1):
   (1/2)(1)^2 + 2(1) = 1/2 + 2 = 5/2
   Subtract: 16 - 5/2 = 27/2
3. Therefore, the definite integral of (x+2) from 1 to 4 is 27/2.

So, ∫(x+2)dx from 1 to 4 equals 27/2.