Question Statement

Q7. Use appropriate formula from geometry to evaluate integrals:

(i) ${\int }_{-1}^4(3-x)dx$

(ii) ${\int }_0^1\left[2+\sqrt{1-x^2}\right]dx$

(iii) ${\int }_2^2\sqrt{x^3-4}dx$

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Background and Explanation

Definite integrals represent the net area between a curve and the $x$-axis over a given interval. When the function graph forms simple geometric shapes (lines, circles, etc.), we can evaluate the integral using basic area formulas from geometry rather than direct integration. This requires identifying the graph, determining the relevant geometric regions, and applying area formulas for triangles, rectangles, and circles.

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Solution

Part (i): ${\int }_{-1}^4(3-x)dx$

The function $y=3-x$ is a linear equation, so its graph is a straight line with slope $-1$.

First, identify key points on the line:

• At $x=-1$: $y=4$, so the point is $(-1,4)$
• At $x=4$: $y=-1$, so the point is $(4,-1)$
• At $x=3$: $y=0$ (where the line crosses the $x$-axis)

The line crosses the $x$-axis at $x=3$, dividing the interval $[-1,4]$ into two regions:

Region from $-1$ to $3$ (above $x$-axis): This forms a right triangle with base $=4$ and height $=4$.

Region from $3$ to $4$ (below $x$-axis): This forms a right triangle with base $=1$ and height $=1$.

Calculating the total area:

$\begin{array}{rl}\text{Required Area}& =\text{Area of }△ACB+\text{Area of }△CDE\\ & =\frac{1}{2}(4)(4)+\frac{1}{2}(1)(1)\\ & =8+\frac{1}{2}\\ & =\frac{17}{2}=8.5\text{ sq. units}\end{array}$

Check by Direct Integration:

Since $y=3-x>0$ for $-1<x<3$ and $y=3-x<0$ for $3<x<4$, we split the integral:

$\text{Area}={\int }_{-1}^3(3-x)dx+\left(-{\int }_3^4(3-x)dx\right)$

Evaluating step-by-step:

$\begin{array}{rl}& =3{\int }_{-1}^{3}1dx-{\int }_{-1}^{3}xdx-3{\int }_3^{4}1dx+{\int }_3^{4}xdx\\ & =3{\left[x\right]}_{-1}^3-{\left[\frac{x^2}{2}\right]}_{-1}^3-3{\left[x\right]}_3^4+{\left[\frac{x^2}{2}\right]}_3^4\\ & =3(3-(-1))-\frac{1}{2}\left((3{)}^2-(-1{)}^2\right)-3(4-3)+\frac{1}{2}\left((4{)}^2-(3{)}^2\right)\\ & =3(4)-\frac{1}{2}(9-1)-3(1)+\frac{1}{2}(16-9)\\ & =12-\frac{1}{2}(8)-3+\frac{1}{2}(7)\\ & =12-4-3+\frac{7}{2}\\ & =5+\frac{7}{2}\\ & =\frac{10+7}{2}\\ & =\frac{17}{2}\\ & =8.5\text{ sq. units}\end{array}$

Part (ii): ${\int }_0^1\left[2+\sqrt{1-x^2}\right]dx$

Split the integral into two separate terms:

${\int }_0^1\left[2+\sqrt{1-x^2}\right]dx={\int }_0^{1}2dx+{\int }_0^1\sqrt{1-x^2}dx$

First term: ${\int }_0^{1}2dx$

Here $y=2$ is a line parallel to the $x$-axis. From the figure, the region from $0$ to $1$ forms a rectangle with base $=1$ and height $=2$.

$\begin{array}{rl}\text{Area}& =(1)\times (2)\\ A_1& =2\end{array}$

Second term: ${\int }_0^1\sqrt{1-x^2}dx$

Here $y=\sqrt{1-x^2}$. Squaring both sides:

$\begin{array}{rl}{y}^2& =1-x^2\\ x^2+y^2& =1\end{array}$

This is the equation of a circle with center $(0,0)$ and radius $=1$. The equation $y=\sqrt{1-x^2}$ represents the upper semi-circle. From the figure, the region from $0$ to $1$ is a quarter circle.

$\begin{array}{rl}\text{Area}& =\frac{1}{4}(\text{Area of circle})\\ A_2& =\frac{1}{4}\left(\pi r^2\right)\\ & =\frac{1}{4}\left(\pi (1{)}^2\right)\\ & =\frac{\pi }{4}\end{array}$

Total Area:

$\begin{array}{rl}\text{Required Area}& =A_1+A_2\\ & =2+\frac{\pi }{4}\end{array}$

Check by Direct Integration:

$\begin{array}{rl}\text{Area}& ={\int }_0^1\left(2+\sqrt{1+x^2}\right)dx\\ & =2{\int }_0^{1}1dx+{\int }_0^1\sqrt{1-x^2}dx\\ & =2{\left[x\right]}_0^1+{\int }_0^1\sqrt{(1{)}^2-(x{)}^2}dx\\ & =2(1-0)+{\left[\frac{x\sqrt{1-x^2}}{2}+\frac{1}{2}{\sin }^{-1}(x)\right]}_0^1\end{array}$

(Formula used: $\int \sqrt{a^2-x^2}dx=\frac{x\sqrt{a^2-x^2}}{2}+\frac{a^2}{2}{\sin }^{-1}\left(\frac{x}{a}\right)$)

$\begin{array}{rl}& =2(1)+\left[\left(\frac{1\sqrt{1-1}}{2}+\frac{1}{2}{\sin }^{-1}(1)\right)-\left(\frac{0\sqrt{1-0}}{2}+\frac{1}{2}{\sin }^{-1}(0)\right)\right]\\ & =2+\left[\left(0+\frac{1}{2}\left(\frac{\pi }{2}\right)\right)-\left(0+\frac{1}{2}(0)\right)\right]\\ & =2+\left[0+\frac{\pi }{4}-(0+0)\right]\\ & =2+\frac{\pi }{4}\end{array}$

Part (iii): ${\int }_2^2\sqrt{x^3-4}dx$

As the integral represents the area under the curve, and if the interval has zero width (both limits are identical), then the area is zero.

$\text{Area}=0$

Check:

$\text{Area}={\int }_2^2\sqrt{x^3-4}dx$

Since both the upper and lower limits are the same, by the property of definite integrals:

${\int }_a^{a}f(x)dx=0$

Therefore:

$\text{Area}=0$

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Key Formulas or Methods Used

• Area of Triangle: $\frac{1}{2}\times \text{base}\times \text{height}$
• Area of Rectangle: $\text{length}\times \text{width}$
• Area of Circle: $\pi r^2$ (quarter circle: $\frac{\pi r^2}{4}$)
• Geometric Interpretation of Definite Integral: ${\int }_a^{b}f(x)dx$ equals net area between curve and $x$-axis
• Integration Formula: $\int \sqrt{a^2-x^2}dx=\frac{x\sqrt{a^2-x^2}}{2}+\frac{a^2}{2}{\sin }^{-1}\left(\frac{x}{a}\right)+C$
• Zero Interval Property: ${\int }_a^{a}f(x)dx=0$

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Summary of Steps

1. Identify the graph: Determine what geometric shape the function represents (line, circle, etc.) over the given interval.
2. Find key points: Calculate intercepts and boundaries to determine dimensions of geometric regions.
3. Split if necessary: If the function crosses the $x$-axis or changes behavior within the interval, break the integral into separate geometric regions.
4. Apply area formulas: Use standard geometric formulas (triangle, rectangle, circle areas) to calculate each component.
5. Sum the areas: Add the individual areas to obtain the total value of the integral.
6. Verify (optional): Confirm the result using direct integration techniques and fundamental theorem of calculus.