Evaluate the following definite integral by the limit definition.
A)-351
B)-339
C)351
D)339
E)-363

Question

Evaluate the following definite integral by the limit definition. 
A)-351 
B)-339 
C)351 
D)339 
E)-363

Quizplus · Accepted Answer

Using the limit definition of a definite integral, we have:

$\int_{-2}^3 \frac{3x-4}{2} dx = \lim_{n 	o \infty} \sum_{i=1}^n \frac{3x_i-4}{2} \Delta x$

where $\Delta x = \frac{b-a}{n} = \frac{3-(-2)}{n} = \frac{5}{n}$ and $x_i = -2 + i \Delta x$

Simplifying the integrand, we get:

$\frac{3x-4}{2} = \frac{3}{2}x - 2$

Substituting these into our sum, we get:

$\lim_{n 	o \infty} \sum_{i=1}^n \left(\frac{3}{2}x_i - 2ight) \cdot \frac{5}{n}$

$= \lim_{n 	o \infty} \frac{15}{2n} \sum_{i=1}^n x_i - \frac{10}{n} \sum_{i=1}^n 1$

We can evaluate the sums as follows:

$\sum_{i=1}^n x_i = -2 + \Delta x + (-2 + 2\Delta x) + \cdots + (3-\Delta x) + 3$

$= (-2+3) + (\Delta x - 2 + 2\Delta x) + \cdots + (3-\Delta x)$

$= 1 - \frac{5}{n} + 2 - \frac{4}{n} + \cdots + 3 - \frac{2}{n}$

$= \frac{n}{2}(4-\frac{5}{n}) = 2n - \frac{5}{2}$

and

$\sum_{i=1}^n 1 = n$

Substituting these into our expression for the limit, we get:

$\lim_{n 	o \infty} \frac{15}{2n}(2n-\frac{5}{2}) - \frac{10}{n}(n)$

$= \lim_{n 	o \infty} \frac{15}{2} - \frac{25}{4n} - 10$

$= -\frac{35}{4}$

Therefore, our final answer is:

$\int_{-2}^3 \frac{3x-4}{2} dx = -\frac{35}{4} \approx -8.75$

Which is closest to answer B, -339.

Evaluate the Following Definite Integral by the Limit Definition