Question Statement

Find the area of the region bounded by the curve $y = x^{2}$ , the $x$ -axis, and the lines $x = 1$ and $x = 3$ .

Background and Explanation

This problem requires calculating the area under a curve using definite integration. The area bounded by a curve $y = f (x)$ , the $x$ -axis, and vertical lines $x = a$ and $x = b$ is given by the definite integral $\int_{a}^{b} f (x) d x$ .

Solution

To find the area of the specified region, we set up a definite integral with the given function $y = x^{2}$ and integrate from $x = 1$ to $x = 3$ .

First, we express the area as the integral of the function with respect to $x$ over the interval $[1, 3]$ :

$Area = \int_{1}^{3} y d x = \int_{1}^{3} x^{2} d x$

Next, we find the antiderivative of $x^{2}$ using the power rule for integration. The antiderivative of $x^{2}$ is $\frac{x ^{3}}{3}$ . We evaluate this at the upper and lower limits according to the Fundamental Theorem of Calculus:

$Area = \frac{x ^{3}}{3}_{1}^{3} = \frac{1}{3} ((3)^{3} - (1)^{3}) = \frac{1}{3} (27 - 1) = \frac{1}{3} (26) = \frac{26}{3} sq. unit$

Therefore, the area of the region bounded by the curve $y = x^{2}$ , the $x$ -axis, and the lines $x = 1$ and $x = 3$ is $\frac{26}{3}$ square units.

Key Formulas or Methods Used

Area under a curve: $Area = \int_{a}^{b} y d x = \int_{a}^{b} f (x) d x$
Power rule for integration: $\int x^{n} d x = \frac{x ^{n + 1}}{n + 1} + C$ (for $n \neq = - 1$ )
Fundamental Theorem of Calculus: $\int_{a}^{b} f (x) d x = [F (x)]_{a}^{b} = F (b) - F (a)$ , where $F (x)$ is the antiderivative of $f (x)$

Summary of Steps

Set up the definite integral $\int_{1}^{3} x^{2} d x$ representing the area under $y = x^{2}$ from $x = 1$ to $x = 3$ .
Find the antiderivative of $x^{2}$ , which is $\frac{x ^{3}}{3}$ .
Apply the Fundamental Theorem of Calculus by evaluating $\frac{x ^{3}}{3}$ at the upper limit ( $x = 3$ ) and lower limit ( $x = 1$ ).
Calculate the difference: $\frac{1}{3} (27 - 1) = \frac{26}{3}$ .
State the final answer with appropriate units: $\frac{26}{3}$ sq. units.

Solution

To find the area of the specified region, we set up a definite integral with the given function

y = x^{2}

and integrate from

x = 1

x = 3

First, we express the area as the integral of the function with respect to

x

over the interval

[1, 3]

Area = \int_{1}^{3} y d x = \int_{1}^{3} x^{2} d x

Next, we find the antiderivative of

x^{2}

using the power rule for integration. The antiderivative of

x^{2}

\frac{x ^{3}}{3}

. We evaluate this at the upper and lower limits according to the Fundamental Theorem of Calculus:

Area = \frac{x ^{3}}{3}_{1}^{3} = \frac{1}{3} ((3)^{3} - (1)^{3}) = \frac{1}{3} (27 - 1) = \frac{1}{3} (26) = \frac{26}{3} sq. unit

Therefore, the area of the region bounded by the curve

y = x^{2}

, the

x

-axis, and the lines

x = 1

and

x = 3

\frac{26}{3}

square units.

Summary of Steps

Set up the definite integral

\int_{1}^{3} x^{2} d x

representing the area under

y = x^{2}

from

x = 1

x = 3

Find the antiderivative of

x^{2}

, which is

\frac{x ^{3}}{3}

Apply the Fundamental Theorem of Calculus by evaluating

\frac{x ^{3}}{3}

at the upper limit (

x = 3

) and lower limit (

x = 1

Calculate the difference:

\frac{1}{3} (27 - 1) = \frac{26}{3}

State the final answer with appropriate units:

\frac{26}{3}

sq. units.

3.8 Q-1

Question Statement

Background and Explanation

Solution

Key Formulas or Methods Used

Summary of Steps

3.8 Q-1

Question Statement

Background and Explanation

Solution

Key Formulas or Methods Used

Summary of Steps