Question

Using the limit process, find the area of the region between the graph $y = {x^2} + 1$ and the x-axis over the interval $[0,3]$?

Answer 1

Hint: We can calculate the area under the curve using thin rectangles. The accuracy of the approximation depends on the number of rectangles we use to approximate. We know that calculus deals with the infinite limit of a finite series of infinitesimally thin rectangles. Here we will use the concept of breaking the interval into subinterval and use the standard summation formula.

Definition of integration: we know that integrating a function between an interval $\int\limits_n^m {f\left( x \right)} dx$ gives the area under the curve between $x = n$ and $x = m$.
Standard summation formula: We use standard summation formula to find the area.
$
  \sum\limits_{r = 1}^n {a = an} \\
  \sum\limits_{r = 1}^n {{r^2}} = \dfrac{1}{6}n\left( {n + 1} \right)\left( {2n + 1} \right) \\
 $

Complete step by step solution:
Use thin rectangles method to calculate the area.
$\int\limits_n^m {f\left( x \right)} = \mathop {\lim }\limits_{C \to \infty } \dfrac{{m - n}}{c}\sum\limits_{i = 1}^c {f\left( {n + i\dfrac{{m - n}}{c}} \right)} $
Break the interval $\left[ {0,3} \right]$ by using equally spacing.
$
  \Delta = \left\{ {n + 0\left( {\dfrac{{m - n}}{c}} \right),n + 1\left( {\dfrac{{m - n}}{c}} \right),.......,n + c\left( {\dfrac{{m - n}}{c}} \right)} \right\} \\
   \Rightarrow \Delta = \left\{ {n,n + 1\left( {\dfrac{{m - n}}{c}} \right),n + 2\left( {\dfrac{{m - n}}{c}} \right),....m} \right\} \\
 $
Substitute $\left( {n = 0} \right)$and $\left( {m = 3} \right)$ in the formula and solve.

By definition of integral we have, $\int\limits_0^3 {\left( {{x^2} + 1} \right)} dx$ so,
$
  \int\limits_0^3 {\left( {{x^2} + 1} \right)} dx = \mathop {\lim }\limits_{c \to \infty } \dfrac{3}{c}\sum\limits_{i = 1}^c {f\left( {0 + i.\dfrac{3}{n}} \right)} \\
   = \mathop {\lim }\limits_{c \to \infty } \dfrac{3}{c}\sum\limits_{i = 1}^c {f\left( {\dfrac{{3i}}{c}} \right)} \\
   = \mathop {\lim }\limits_{c \to \infty } \dfrac{3}{c}\sum\limits_{i = 1}^c {\left\{ {{{\left( {\dfrac{{3i}}{c}} \right)}^2} + 1} \right\}} \\
   = \mathop {\lim }\limits_{c \to \infty } \dfrac{3}{c}\sum\limits_{i = 1}^c {\left\{ {\dfrac{{9{i^2}}}{{{c^2}}} + \sum\limits_{i = 1}^c 1 } \right\}} \\
 $

Using the standard summation formula we have,
$
  I = \mathop {\lim }\limits_{c \to \infty } \dfrac{3}{c}\left\{ {\dfrac{3}{{2{c^2}}}\left( {2{c^2} + 3c + 1} \right) + 1} \right\} \\
   \Rightarrow I = \mathop {\lim }\limits_{c \to \infty } \dfrac{3}{{2{c^2}}}\left\{ {8{c^2} + 9c + 3} \right\} \\
   \Rightarrow I = \mathop {\lim }\limits_{c \to \infty } \dfrac{{24{c^2} + 27c + 9}}{{2{c^2}}} \\
 $
Now substitute the limit in the equation to find the area under the curve between [0,3].
$
  I = \mathop {\lim }\limits_{c \to \infty } \dfrac{{24{c^2} + 27c + 9}}{{2{c^2}}} \\
   \Rightarrow I = \left( {12 + \dfrac{{27}}{{2c}} + \dfrac{9}{{2{c^2}}}} \right) \\
   \Rightarrow I = 12 \\
 $

Therefore the area under the curve by using the process of limit is calculated and equals to 12.

Note: To find the area under the curve use the thin rectangles method, By using equal spacing formula to break the interval. The formula of breaking the interval is $\Delta = \left\{ {n + 0\left( {\dfrac{{m - n}}{c}} \right),n + 1\left( {\dfrac{{m - n}}{c}} \right),.......,n + c\left( {\dfrac{{m - n}}{c}} \right)} \right\}$.
Use the definition of integral formula $\left( {\int\limits_n^m {f\left( x \right)} = \mathop {\lim }\limits_{C \to \infty } \dfrac{{m - n}}{c}\sum\limits_{i = 1}^c {f\left( {n + i\dfrac{{m - n}}{c}} \right)} } \right)$ to calculate the area.