Express the integral as a limit of Riemann sums?

Question

#\int_{4}^{12}[\ln(1+x^2)-\sin(x)]dx#
Do not evaluate the limit.

Please don't use anything above Calculus I level.

Ezekiel Alexander · Accepted Answer

Here is a limit definition of the definite integral. (I'd guess 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 go one small step at a time.

#int_4^12 [ln(1+x^2)-sinx] dx#.

Find #Delta x#

For each #n#, we get

#Deltax = (b-a)/n = (12-4)/n = 8/n#

Find #x_i#

And #x_i = a+iDeltax = 4+i8/n = 4+(8i)/n#

Find #f(x_i)#

#f(x_i) = ln(1+x_i""^2)-sinx_i = ln(1+(4+(8i)/n)^2)-sin(4+(8i)/n)#

#int_4^12 [ln(1+x^2)-sinx] dx#

# = lim_(nrarroo)sum_(i=1)^n [ (ln(1+(4+(8i)/n)^2)-sin(4+(8i)/n))8/n]#.

Expand/simplify #(1+(4+(8i)/n)^2)# if required to #(17+(64i)/n + (64i^2)/n^2)#

Charles Anctil · Answer

undefined To express an integral as a limit of Riemann sums, you use the following formula:

$$
\int_{a}^{b} f(x) \, dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*) \cdot \Delta x
$$

where:
- $a$ and $b$ are the lower and upper bounds of integration respectively,
- $f(x)$ is the function being integrated,
- $n$ is the number of partitions,
- $x_i^*$ is a sample point in the $i$th subinterval,
- $\Delta x$ is the width of each subinterval.

The Riemann sum is formed by summing the products of the function evaluated at the sample points $x_i^*$ and the width of each subinterval $\Delta x$, and as $n$ approaches infinity, the Riemann sum approaches the integral.

Express the integral as a limit of Riemann sums?

#\int_{4}^{12}[\ln(1+x^2)-\sin(x)]dx# Do not evaluate the limit. Please don't use anything above Calculus I level.

#\int_{4}^{12}[\ln(1+x^2)-\sin(x)]dx#
Do not evaluate the limit.

Please don't use anything above Calculus I level.