Question Statement

Sketch the region where area is represented by the definite integral and evaluate the integral using an appropriate formula from geometry.

(i) ${\int }_0^{4}xdx$

(ii) ${\int }_{-3}^{0}xdx$

(iii) ${\int }_0^2(x-1)dx$

(iv) ${\int }_0^2(x+1)dx$

(v) ${\int }_{-3}^{3}2dx$

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Background and Explanation

Definite integrals calculate the net area between a curve $y=f(x)$ and the $x$-axis over an interval $[a,b]$. When $f(x)$ is a linear function, the bounded region forms simple geometric shapes (triangles, rectangles, or trapezoids), allowing us to verify integration results using basic area formulas from geometry.

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Solution

Part (i): ${\int }_0^{4}xdx$

Given $y=x$, which is a linear equation representing a straight line passing through the origin.

Find the boundary points:

• At $x=0$, $y=0$. Point is $(0,0)$
• At $x=4$, $y=4$. Point is $(4,4)$

From the figure, the region is a right-angled triangle with:

• Base $=4$
• Height $=4$

Using the area formula for a triangle: $\text{Area}=\frac{1}{2}\times \text{base}\times \text{height}=\frac{1}{2}\times 4\times 4=8\text{ sq. units}$

Verification by integration: $\begin{array}{rl}\text{Area}& ={\int }_0^{4}xdx\\ & ={\left[\frac{x^2}{2}\right]}_0^4\\ & =\frac{1}{2}{\left[x^2\right]}_0^4\\ & =\frac{1}{2}\left[(4{)}^2-(0{)}^2\right]\\ & =\frac{1}{2}(16-0)\\ & =8\text{ sq. units}\end{array}$

Part (ii): ${\int }_{-3}^{0}xdx$

Given $y=x$, a straight line.

Find the boundary points:

• At $x=0$, $y=0$. Point is $(0,0)$
• At $x=-3$, $y=-3$. Point is $(-3,-3)$

From the figure, the region is a right-angled triangle in the third quadrant with:

• Base $=3$ (absolute value of horizontal distance from $-3$ to $0$)
• Height $=3$ (absolute value of vertical distance)

Using the geometric formula: $\text{Area}=\frac{1}{2}\times 3\times 3=\frac{9}{2}=4.5\text{ sq. units}$

Verification by integration: $\begin{array}{rl}\text{Area}& ={\int }_{-3}^{0}xdx\\ & ={\left[\frac{x^2}{2}\right]}_{-3}^0\\ & =\frac{1}{2}{\left[x^2\right]}_{-3}^0\\ & =\frac{1}{2}\left[(0{)}^2-(-3{)}^2\right]\\ & =\frac{1}{2}(0-9)\\ & =-\frac{9}{2}\end{array}$

Since area is always positive, we take the absolute value: $\text{Area}=\frac{9}{2}=4.5\text{ sq. units}$

Part (iii): ${\int }_0^2(x-1)dx$

Given $y=x-1$, a linear equation.

Find the boundary points:

• At $x=0$, $y=-1$. Point is $(0,-1)$
• At $x=2$, $y=1$. Point is $(2,1)$

The line crosses the $x$-axis at $x=1$ (where $y=0$).

From the figure, the region consists of two right-angled triangles:

• Triangle below $x$-axis from $x=0$ to $x=1$ (base 1, height 1)
• Triangle above $x$-axis from $x=1$ to $x=2$ (base 1, height 1)

Total area is the sum of absolute areas: $\text{Area}=\text{Area of }△OAB+\text{Area of }△BCD=\frac{1}{2}(1)(1)+\frac{1}{2}(1)(1)=\frac{1}{2}+\frac{1}{2}=1\text{ sq. unit}$

Verification by integration: Since $y=x-1<0$ for $0<x<1$ and $y=x-1>0$ for $1<x<2$, we split the integral: $\begin{array}{rl}\text{Area}& =-{\int }_0^1(x-1)dx+{\int }_1^2(x-1)dx\left(\text{Area below }x\text{-axis}=-{\int }_a^{b}f(x)dx\right)\\ & =-{\int }_0^{1}xdx+{\int }_0^{1}1dx+{\int }_1^{2}xdx-{\int }_1^{2}1dx\\ & =-{\left[\frac{x^2}{2}\right]}_0^1+{\left[x\right]}_0^1+{\left[\frac{x^2}{2}\right]}_1^2-{\left[x\right]}_1^2\\ & =-\frac{1}{2}(1-0)+(1-0)+\frac{1}{2}(4-1)-(2-1)\\ & =-\frac{1}{2}+1+\frac{3}{2}-1\\ & =-\frac{1}{2}+\frac{3}{2}\\ & =1\text{ sq. unit}\end{array}$

Part (iv): ${\int }_0^2(x+1)dx$

Given $y=x+1$, a straight line.

Find the boundary points:

• At $x=0$, $y=1$. Point is $(0,1)$
• At $x=2$, $y=3$. Point is $(2,3)$

From the figure, the region is a trapezium with:

• Parallel sides (heights at $x=0$ and $x=2$): $1$ and $3$
• Distance between parallel sides (width): $2$

Using the area formula for a trapezium: $\text{Area}=\frac{1}{2}\times (\text{Sum of parallel sides})\times (\text{Distance between them})=\frac{1}{2}(1+3)(2)=\frac{1}{2}(4)(2)=4\text{ sq. units}$

Verification by integration: $\begin{array}{rl}\text{Area}& ={\int }_0^2(x+1)dx\\ & ={\int }_0^{2}xdx+{\int }_0^{2}1dx\\ & ={\left[\frac{x^2}{2}\right]}_0^2+{\left[x\right]}_0^2\\ & =\frac{1}{2}(4-0)+(2-0)\\ & =2+2\\ & =4\text{ sq. units}\end{array}$

Part (v): ${\int }_{-3}^{3}2dx$

Given $y=2$, which represents a horizontal line parallel to the $x$-axis.

From the figure, the region is a rectangle with:

• Length (along $x$-axis): $3-(-3)=6$
• Width (height): $2$

Using the area formula for a rectangle: $\text{Area}=\text{Length}\times \text{Width}=6\times 2=12\text{ sq. units}$

Verification by integration: $\begin{array}{rl}\text{Area}& ={\int }_{-3}^{3}2dx\\ & =2{\left[x\right]}_{-3}^3\\ & =2(3-(-3))\\ & =2(3+3)\\ & =2(6)\\ & =12\text{ sq. units}\end{array}$

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Key Formulas or Methods Used

• Area of Triangle: $A=\frac{1}{2}\times \text{base}\times \text{height}$
• Area of Trapezium: $A=\frac{1}{2}\times (\text{sum of parallel sides})\times \text{distance between them}$
• Area of Rectangle: $A=\text{length}\times \text{width}$
• Fundamental Theorem of Calculus: ${\int }_a^{b}f(x)dx=F(b)-F(a)$, where $F$ is the antiderivative of $f$
• Signed Area Property: For regions below the $x$-axis, the definite integral yields a negative value; geometric area requires taking the absolute value or adjusting the integral limits with a negative sign: $\text{Area}=-{\int }_a^{b}f(x)dx$ when $f(x)<0$ on $(a,b)$

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Summary of Steps

1. Part (i): Identify line $y=x$ from $(0,0)$ to $(4,4)$; calculate area of right triangle with base 4, height 4: $\frac{1}{2}(4)(4)=8$.
2. Part (ii): Identify line $y=x$ from $(-3,-3)$ to $(0,0)$; calculate area of right triangle with base 3, height 3: $\frac{1}{2}(3)(3)=4.5$ (take absolute value of negative integral result).
3. Part (iii): Identify line $y=x-1$ from $(0,-1)$ to $(2,1)$ crossing at $(1,0)$; sum areas of two triangles (each $\frac{1}{2}\times 1\times 1$) or split integral at $x=1$ with sign adjustment: $1$.
4. Part (iv): Identify line $y=x+1$ from $(0,1)$ to $(2,3)$; calculate trapezium area with parallel sides 1 and 3, width 2: $\frac{1}{2}(1+3)(2)=4$.
5. Part (v): Identify horizontal line $y=2$ from $x=-3$ to $x=3$; calculate rectangle area $6\times 2=12$.