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

2 Answers
Apr 17, 2017

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.

Explanation:

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#

Apr 18, 2017

Answer:

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

Explanation:

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#