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

Question

Ezekiel Alexander · Accepted Answer

Here is a limit definition of the definite integral. (I don't know if it's the one you are using.)

.#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_1^4 (x^2+4x-2) dx#.

Find #Delta x#

For each #n#, we get

#Deltax = (b-a)/n = (4-1)/n = 3/n#

Find #x_i#

And #x_i = a+iDeltax = 1+i3/n = 1+(3i)/n#

Find #f(x_i)#

#f(x_i) = (x_i)^2 + 4(x_i) - 2 = (1+(3i)/n)^2 + 4(1+(3i)/n) - 2 #

# = 1+(6i)/n+(9i^2)/n^2 +4+(12i)/n-2#

# = (9i^2)/n^2 + (18i)/n + 3#

Find and simplify #sum_(i=1)^n f(x_i)Deltax # in order to evaluate the sums.

#sum_(i=1)^n f(x_i)Deltax = sum_(i=1)^n( (9i^2)/n^2 + (18i)/n + 3) 3/n#

# = sum_(i=1)^n( (27i^2)/n^3 + (54i)/n^2 + 9/n)#

# =sum_(i=1)^n (27i^2)/n^3 + sum_(i=1)^n(54i)/n^2 + sum_(i=1)^n 9/n #

# =27/n^3sum_(i=1)^n i^2 + 54/n^2 sum_(i=1)^n i + 9/n sum_(i=1)^n 1 #

Evaluate the sums

# = 27/n^3((n(n+1)(2n+1))/6) + 45/n^2((n(n+1))/2) + 15/n(n) #

(We used summation formulas for the sums in this step.)

Rewrite before finding the limit

#sum_(i=1)^n f(x_i)Deltax = 27/n^3((n(n+1)(2n+1))/6) + 54/n^2((n(n+1))/2) + 9/n(n) #

# = 27/6((n(n+1)(2n+1))/n^3) + 54/2((n(n+1))/n^2) + 9 #

# = 9/2((n(n+1)(2n+1))/n^3) + 27((n(n+1))/n^2) + 9 #

Now we need to evaluate the limit as #nrarroo#.

#lim_(nrarroo) ((n(n+1)(2n+1))/n^3) = 2#

#lim_(nrarroo) ((n(n+1))/n^2) = 1#

To finish the calculation, we have

#int_1^4 (x^2+4x-2) dx= lim_(nrarroo) (9/2((n(n+1)(2n+1))/n^3) + 27((n(n+1))/n^2) + 9)#

# = 9/2(2) + 27(1) + 9 #

# = 45#

Cameron Alexander · Answer

undefined To evaluate the integral $ \int_{1}^{4} (x^2 + 4x - 2) \, dx $ using limits, you first find the antiderivative of the function $ x^2 + 4x - 2 $, which is $ \frac{1}{3}x^3 + 2x^2 - 2x $. Then, you evaluate this antiderivative at the upper limit (4) and subtract the result of evaluating it at the lower limit (1). So, $ \int_{1}^{4} (x^2 + 4x - 2) \, dx = \left[ \frac{1}{3} \cdot 4^3 + 2 \cdot 4^2 - 2 \cdot 4 \right] - \left[ \frac{1}{3} \cdot 1^3 + 2 \cdot 1^2 - 2 \cdot 1 \right] $. Calculating this expression gives the value of the definite integral.

Paisley Johnson · Answer

undefined To evaluate the definite integral $ \int_{1}^{4} (x^2 + 4x - 2) \, dx $ using limits, follow these steps:

1. First, find the antiderivative of the integrand, which is $ \frac{1}{3}x^3 + 2x^2 - 2x $.

2. Evaluate the antiderivative at the upper and lower limits of integration:
   - At $ x = 4 $: $ \frac{1}{3}(4)^3 + 2(4)^2 - 2(4) = \frac{64}{3} + 32 - 8 = \frac{64}{3} + 24 = \frac{112}{3} $
   - At $ x = 1 $: $ \frac{1}{3}(1)^3 + 2(1)^2 - 2(1) = \frac{1}{3} + 2 - 2 = \frac{1}{3} $

3. Subtract the value of the antiderivative at the lower limit from the value at the upper limit:
   $ \frac{112}{3} - \frac{1}{3} = \frac{111}{3} = 37 $

Therefore, $ \int_{1}^{4} (x^2 + 4x - 2) \, dx = 37 $.