Question

How do you find the area under the curve $f\left( x \right)={{x}^{2}}$ on the interval $\left[ -1,1 \right]$?

Answer 1

Hint: Now to find the area under the curve we will integrate the function from -1 to 1.
Now we know that the integration of ${{x}^{n}}=\dfrac{{{x}^{n+1}}}{n+1}$ . Hence we will use this formula to find the antiderivative of ${{x}^{2}}$ and then find the definite integral. Hence we get the area of the given function in the interval $\left[ -1,1 \right]$ .

Complete step by step solution:
Now we are given the function $f\left( x \right)={{x}^{2}}$ .
Now the function is continuous in the interval $\left[ -1,1 \right]$
To find the area under the curve we will use the concept of definite integrals.
Now we know that the area of function $f\left( x \right)$ in $\left[ a,b \right]$ is given by the definite integral $\int_{a}^{b}{f\left( x \right)dx}$ .
Hence now we have $f\left( x \right)={{x}^{2}}$ and interval as $\left[ -1,1 \right]$
Hence we get a = - 1 and b = 1.
Now substituting the values of function f, a and b we get the area as,
$\Rightarrow A=\int_{-1}^{1}{{{x}^{2}}dx}$
Now we will solve the definite integral using the fundamental theorem of calculus.
Now we know that integration of $\int{{{x}^{n}}}=\dfrac{{{x}^{n+1}}}{n+1}+C$ hence using this we get the integration as,
$\begin{align}
  & \Rightarrow A=\left[ \dfrac{{{x}^{3}}}{3} \right]_{-1}^{1} \\
 & \Rightarrow A=\left[ \dfrac{1}{3}-\left( \dfrac{1}{3} \right) \right] \\
 & \Rightarrow A=\dfrac{2}{3} \\
\end{align}$
Hence we get the area under the given curve is $\dfrac{2}{3}$ square units.

Note: Now note that to integrate the function from a to b we divide the area of the function in very small n rectangles of the height $f\left( x \right)$ and with $\dfrac{b-a}{n}$ . Now we take summation of all these areas and hence we can say that the area under the curve is nothing but the integration of the function from a to b. Hence the area under the curve is given by $\int_{a}^{b}{f\left( x \right)dx}$ .