Question Statement

In the figure, a sketch of the function $y = \frac{1}{2} (0.2 x^{2} + x)$ is shown. Find:

(i) the area of region $A$ (from $x = 1$ to $x = 3$ ).

(ii) the area of region $B$ (from $x = 3$ to $x = 4$ ).

(iii) the area of the region from $x = 1$ to $x = 4$ .

(iv) the area of the region from $x = - 1$ to $x = - 4$ .

Background and Explanation

The area bounded by a curve $y = f (x)$ , the $x$ -axis, and the vertical lines $x = a$ and $x = b$ is calculated using the definite integral $\int_{a}^{b} y d x$ . This requires applying the power rule for integration, $\int x^{n} d x = \frac{x ^{n + 1}}{n + 1}$ , and evaluating the antiderivative at the bounds using the Fundamental Theorem of Calculus.

Solution

Part (i): Area of region $A$

Region $A$ corresponds to the area under the curve from $x = 1$ to $x = 3$ . We set up the definite integral:

$Area = \int_{1}^{3} y d x = \int_{1}^{3} \frac{1}{2} (0.2 x^{2} + x) d x$

Factor out the constant $\frac{1}{2}$ and split the integral into two terms:

$= \frac{1}{2} (0.2) \int_{1}^{3} x^{2} d x + \frac{1}{2} \int_{1}^{3} x d x$

$= \frac{1}{10} \int_{1}^{3} x^{2} d x + \frac{1}{2} \int_{1}^{3} x d x$

Apply the power rule for integration to find the antiderivatives:

$= \frac{1}{10} [\frac{x ^{3}}{3}]_{1}^{3} + \frac{1}{2} [\frac{x ^{2}}{2}]_{1}^{3}$

Evaluate at the upper bound ( $x = 3$ ) and lower bound ( $x = 1$ ):

$= \frac{1}{30} (3^{3} - 1^{3}) + \frac{1}{4} (3^{2} - 1^{2})$

$= \frac{1}{30} (27 - 1) + \frac{1}{4} (9 - 1)$

$= \frac{26}{30} + \frac{8}{4}$

Simplify the fractions:

$= \frac{13}{15} + 2 = \frac{13}{15} + \frac{30}{15} = \frac{43}{15} \approx 2.87 sq. units$

Part (ii): Area of region $B$

Region $B$ extends from $x = 3$ to $x = 4$ . Using the same antiderivatives:

$Area = \int_{3}^{4} \frac{1}{2} (0.2 x^{2} + x) d x$

$= \frac{1}{10} [\frac{x ^{3}}{3}]_{3}^{4} + \frac{1}{4} [\frac{x ^{2}}{2}]_{3}^{4}$

$= \frac{1}{30} (4^{3} - 3^{3}) + \frac{1}{4} (4^{2} - 3^{2})$

$= \frac{1}{30} (64 - 27) + \frac{1}{4} (16 - 9)$

$= \frac{37}{30} + \frac{7}{4}$

Converting to decimal form for the final sum:

$= 1.233... + 1.75 = 2.983 sq. units$

(Exact value: $\frac{179}{60}$ sq. units)

Part (iii): Area from $x = 1$ to $x = 4$

This represents the total area covering both regions $A$ and $B$ . By the additive property of definite integrals:

$Area = \int_{1}^{4} y d x = \int_{1}^{3} y d x + \int_{3}^{4} y d x$

Substituting the values from parts (i) and (ii):

$= 2.87 + 2.983 = 5.853 sq. units$

Part (iv): Area from $x = - 1$ to $x = - 4$

Note: When calculating geometric area, we integrate from the lower limit to the upper limit. Here we integrate from $x = - 4$ to $x = - 1$ (since $- 4 < - 1$ ). If the function lies below the $x$ -axis in this interval, the integral will be negative, and we take the absolute value for the physical area.

$Area = \int_{- 4}^{- 1} \frac{1}{2} (0.2 x^{2} + x) d x$

$= \frac{1}{10} [\frac{x ^{3}}{3}]_{- 4}^{- 1} + \frac{1}{4} [\frac{x ^{2}}{2}]_{- 4}^{- 1}$

$= \frac{1}{30} ((- 1)^{3} - (- 4)^{3}) + \frac{1}{4} ((- 1)^{2} - (- 4)^{2})$

$= \frac{1}{30} (- 1 - (- 64)) + \frac{1}{4} (1 - 16)$

$= \frac{1}{30} (63) + \frac{1}{4} (- 15)$

$= 2.1 - 3.75 = - 1.65$

Since area is always positive, we take the absolute value:

$∴ Area = 1.65 sq. units$

Key Formulas or Methods Used

Definite Integral for Area: $Area = \int_{a}^{b} f (x) d x$
Power Rule: $\int x^{n} d x = \frac{x ^{n + 1}}{n + 1} + C$
Fundamental Theorem of Calculus: $\int_{a}^{b} f (x) d x = F (b) - F (a)$ , where $F$ is the antiderivative of $f$
Additive Property: $\int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x$
Geometric Area: Always non-negative; take $∣ integral ∣$ if the region is below the $x$ -axis

Summary of Steps

Set up the integral: Identify the bounds for each region and write $\int_{a}^{b} \frac{1}{2} (0.2 x^{2} + x) d x$ .
Simplify the integrand: Factor out constants to get $\frac{1}{10} x^{2} + \frac{1}{2} x$ .
Integrate term by term: Apply the power rule to obtain antiderivatives $\frac{x ^{3}}{30} + \frac{x ^{2}}{4}$ .
Evaluate bounds: Substitute upper and lower limits using $F (b) - F (a)$ for each part.
Calculate numerical values: Simplify fractions and sum the results.
Handle negative regions: For intervals where the curve is below the $x$ -axis (yielding a negative integral), take the absolute value to report the geometric area.
Combine regions: For part (iii), add the areas of regions $A$ and $B$ obtained previously.