Solveeit Logo

Question

Question: How do you use the limit process to find the area of the region between the graph \(y = {x^2} + 1\) ...

How do you use the limit process to find the area of the region between the graph y=x2+1y = {x^2} + 1 and the xx axis over the interval [0,3]\left[ {0,3} \right] ?

Explanation

Solution

Here we will use integration to find the area of the region bounded on the certain limit. The area AA of the region under the graph of a function f(x)0f(x) \geqslant 0 above the xx axis from x=ax = a and x=bx = b can be found by A=abf(x)dxA = \int\limits_a^b {f(x)\,dx} .
Formula used:
If f(x)=xnf(x) = {x^n} then xndx=xn+1n+1+C\int {{x^n}\,dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C

Complete step by step answer: By the definition of an integral
A=abf(x)dxA = \int\limits_a^b {f(x)\,dx} represents the area under the curve y=f(x)y = f(x) between x=ax = a and x=bx = b.
Now we have the function y=x2+1y = {x^2} + 1 and the interval [0,3]\left[ {0,3} \right]
Let write down in integration form, that is,
A=03x2+1dx\Rightarrow A = \int\limits_0^3 {{x^2} + 1\,dx}
Here the upper limit value is x=0x = 0and lower limit value is x=3x = 3
Now using the integration property,
f(x)+g(x)dx=f(x)dx+g(x)dx\int {f(x) + g(x)\,dx} = \int {f(x)\,dx + \int {g(x)\,dx} }
By the above property separate the integration,
A=03x2dx+031dx\Rightarrow A = \int\limits_0^3 {{x^2}\,dx} + \int\limits_0^3 {1\,dx}
Now using the integration formula that is,
xndx=xn+1n+1+C\int {{x^n}\,dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C
Now we get the result,
A=[x2+12+1]03+[x]03\Rightarrow A = \left[ {\dfrac{{{x^{2 + 1}}}}{{2 + 1}}} \right]_0^3 + \left[ x \right]_0^3
On simplified,
A=[x33]03+[x]03\Rightarrow A = \left[ {\dfrac{{{x^3}}}{3}} \right]_0^3 + \left[ x \right]_0^3
Now applying the upper limit and the lower limit, we know that while applying the limit we must follow the rule that is we first apply the upper limit value to the variable xx and then subtract the resulting value with the lower limit value which is applied to the variable xx.
    \;\; A=[3330]+[30] \Rightarrow A = \left[ {\dfrac{{{3^3}}}{3} - 0} \right] + \left[ {3 - 0} \right]
Simplify the above we get,
A=[32]+[3]\Rightarrow A = \left[ {{3^2}} \right] + \left[ 3 \right]
On square of the required number,
A=9+3\Rightarrow A = 9 + 3
By adding,
A=12\Rightarrow A = 12
Hence the area of the region of a given function bounded within the interval is 1212.

Note:
The area under a curve between two points can be found by doing a definite integral between the two points.
To find the area of the region y=f(x)y = f(x) between x=ax = a and x=bx = b , integrate y=f(x)y = f(x) between the limits of aa and bb.
Areas under the xx axis will come out negative and areas above the xx axis will be positive.
This means that you have to be careful when finding an area which is partly above and partly below the xx axis.