How do you evaluate the definite integral by the limit definition given #int (2x+5)dx# from [0,2]?

Question

William Abdalla · Accepted Answer

Here is a limit definition of the definite integral. (I hope 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.)

Let's do one small step at a time.

#int_0^2 (2x+5) dx#.

Find #Delta x#

For each #n#, we get

#Deltax = (b-a)/n = (2-0)/n = 2/n#

Find #x_i#

And #x_i = a+iDeltax = 0+i2/n = (2i)/n#

Find #f(x_i)#

#f(x_i) = 2(x_i) +5 #

# = 2((2i)/n) +5#

# = (4i)/n+5#

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

#sum_(i=1)^n f(x_i)Deltat = sum_(i=1)^n ( (4i)/n+5) 2/n#

# = sum_(i=1)^n( (8i)/n^2 + 10/n)#

# = 8/n^2 sum_(i=1)^n(i) + 10/n sum_(i=1)^n 1#

Evaluate the sums

# = 8/n^2((n(n+1))/2) + 10/n(n)#

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

Rewrite before finding the limit

#sum_(i=1)^n f(x_i)Deltax = 8/n^2((n(n+1))/2) +10/n(n)#

# = 4((n(n+1))/n^2)) + 10#

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

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

To finish the calculation, we have

#int_0^2 (2x+5) dx = lim_(nrarroo) ( 4((n(n+1))/n^2))+10#

# = 4(1) +10 = 14#

Paisley Johnson · Answer

undefined To evaluate the definite integral $\int_0^2 (2x+5)dx$ using the limit definition, follow these steps:

1. Start with the integral expression: $\int_a^b f(x)dx$.
2. Split the interval $[a, b]$ into $n$ subintervals of equal width: $\Delta x = \frac{b-a}{n}$.
3. Choose $x_i^*$ as the representative point in the $i$th subinterval, where $x_i^*$ lies in the interval $[x_{i-1}, x_i]$.
4. Form the Riemann sum: $R_n = \sum_{i=1}^n f(x_i^*)\Delta x$.
5. Take the limit of $R_n$ as $n$ approaches infinity: $\lim_{n\to\infty} R_n$.
6. This limit represents the value of the definite integral: $\int_a^b f(x)dx$.

For the given integral $\int_0^2 (2x+5)dx$:

1. $a = 0$ and $b = 2$.
2. $f(x) = 2x+5$.
3. Calculate $\Delta x = \frac{2-0}{n} = \frac{2}{n}$.
4. Choose $x_i^* = \frac{2i}{n}$.
5. Form the Riemann sum: 
   $$R_n = \sum_{i=1}^n f(x_i^*)\Delta x$$
   $$= \sum_{i=1}^n \left(2\left(\frac{2i}{n}\right) + 5\right)\frac{2}{n}$$
   $$= \frac{4}{n} \sum_{i=1}^n i + \frac{10}{n} \sum_{i=1}^n 1$$
   $$= \frac{4}{n} \cdot \frac{n(n+1)}{2} + \frac{10}{n} \cdot n$$
   $$= 2(n+1) + 10$$
   $$= 2n + 2 + 10$$
   $$= 2n + 12$$.
6. Take the limit:
   $$\lim_{n\to\infty} R_n = \lim_{n\to\infty} (2n + 12) = \infty.$$

Therefore, the value of the definite integral $\int_0^2 (2x+5)dx$ is infinite.