How do you evaluate the integral of #int (3 - 2x)dx# from -1 to 3?

Question

Emilia Aguilar · Accepted Answer

Here is a geometric approach.

Consider the graph of #y=3-2x#.

The integral is the area under the curve above the #x# axis, minus the area below the #x# axis and above the curve.

In the image below, the integral is the blue area minus the red area.

The blue triangle has base #5/2# and height #5# for an area of #25/4#.

The red triangle has base #3/2# and height #3# for an area of #9/4#

#int_01^3 (3-2x) dx = 25/4 = 9/4 = 4#

Emily Ammerman · Answer

To evaluate the integral of a simple polynomial, simply do the inverse of the power rule for differentiation by adding one to the exponent, and then dividing by that new exponent. Then, find the value of the function by subtracting the function at the smaller bound from the function evaluated at the larger bound.

For your example, integrating we get: #int_-1^3 (3-2x)dx = (3x - x^2)|_-1^3 # Now substitute the bounds in: # (3x - x^2)|_-1^3 = [3(3)-(3)^2] - [3(-1)-(-1)^2] # Simplifying we get: # [0] - [-4] = 4 # Therefore, #int_-1^3 (3-2x)dx# amounts to 4. Hopefully this was clear! If you have any questions, feel free to ask! :)

Paisley Johnson · Answer

If you do not yet have the Fundamental Theorem of Calculus available to evaluate this, but you need to evaluate it from a definition, see below.

.#int_a^b f(x) dx = lim_(nrarroo) sum_(i=1)^n f(x_i)Deltax#. Where, for each positive integer #n#, we let #Deltax = (b-a)/n# And for #i=1,2,3, . . . ,n#, we let #x_i = a+iDeltax#. (These #x_i# are the right endpoints of the subintervals.) I prefer to do this type of problem one small step at a time. #int_-1^3 (3-2x)dx#. For each #n#, we get #Deltax = (b-a)/n = (3-(-1))/n = 4/n# And #x_i = a+iDeltax = -1+i4/n =-1+ (4i)/n# #f(x_i) = 3-2x_i =3-2 (-1+(4i)/n)# # = 5 - (8i)/n# #sum_(i=1)^n f(x_i)Deltax = sum_(i=1)^n(5 - (8i)/n) 4/n# # = sum_(i=1)^n(20/n-(32i)/n^2)# # = 20/n sum_(i=1)^n 1 + 32/n^2 sum_(i=1)^n i # # = 20/n[n] - 32/n^2[(n(n+1))/2]# (We used summation formulas for the sums in the previous step.) So, #sum_(i=1)^n f(x_i)Deltax = 20 - 32/2 [(n(n+1))/n^2]# The last thing to do is evaluate the limit as #nrarroo#. I hope it is clear that this amounts to evaluating #lim_(nrarroo)[(n(n+1))/n^2]# There are several ways to think about this limit : The numerator can be expanded to a polynomial with leading term #n^2#, so the limit as #nrarroo# is #1#. OR #(n(n+1))/n^2 = (n/n)((n+1)/n) = (1)(n/n+1/n) = 1+1/n# The limit at infinity is #1# Completing the integration .#int_-1^3 (3-2x)dx = lim_(nrarroo) sum_(i=1)^n f(x_i)Deltax# # = lim_(nrarroo) sum_(i=1)^n(5 - (8i)/n) 4/n# # = lim_(nrarroo) [20/n[n] - 32/n^2[(n(n+1))/2]]# # = 20-32/2lim_(nrarroo)(n(n+1))/n^2# # = 20-16 = 4#

Ezekiel Alexander · Answer

undefined To evaluate the integral ∫(3 - 2x)dx from -1 to 3, you first need to integrate the expression (3 - 2x) with respect to x and then evaluate it at the upper limit (3) and subtract the value obtained at the lower limit (-1).

∫(3 - 2x)dx = [3x - x^2] evaluated from -1 to 3

Substituting the upper limit (3) into the expression gives:
3(3) - (3)^2 = 9 - 9 = 0

Substituting the lower limit (-1) into the expression gives:
3(-1) - (-1)^2 = -3 - 1 = -4

Therefore, the value of the integral from -1 to 3 is 0 - (-4) = 4.