Hello future Engineers! Welcome to this deep dive into one of the most practical and geometrically intuitive applications of definite integrals:
determining areas of regions bounded by simple curves in standard forms. This topic is not just about crunching numbers; it's about visualizing shapes, understanding their boundaries, and using the power of calculus to measure their extent. For IIT JEE, this is a consistently tested area, demanding both conceptual clarity and strong integration skills.
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The Fundamental Principle: Area as a Definite Integral
Recall from your basics that a
definite integral, $int_a^b f(x) , dx$, geometrically represents the area bounded by the curve $y=f(x)$, the x-axis, and the vertical lines $x=a$ and $x=b$, provided $f(x) ge 0$ in the interval $[a, b]$.
Imagine slicing the area under a curve into an infinite number of extremely thin, vertical rectangular strips. Each strip has a width $dx$ (infinitesimally small) and a height $f(x)$. The area of one such strip is approximately $f(x) cdot dx$. Summing up these infinitesimal areas from $x=a$ to $x=b$ gives us the total area. This is precisely what a definite integral does!
Key Idea: The definite integral is essentially an "infinite sum" of infinitesimally small parts. For area, these parts are tiny rectangular strips.
Mathematically:
1.
Area with respect to the x-axis: If a region is bounded by $y = f(x)$, the x-axis, and the lines $x=a$ and $x=b$, the area is given by:
$A = int_a^b |f(x)| , dx$
We use the absolute value because area is always non-negative. If $f(x)$ is entirely above the x-axis ($f(x) ge 0$), then $A = int_a^b f(x) , dx$. If it's entirely below the x-axis ($f(x) le 0$), then $A = int_a^b -f(x) , dx$.
2.
Area with respect to the y-axis: Similarly, if a region is bounded by $x = g(y)$, the y-axis, and the lines $y=c$ and $y=d$, the area is:
$A = int_c^d |g(y)| , dy$
Again, we use the absolute value. If $g(y)$ is entirely to the right of the y-axis ($g(y) ge 0$), then $A = int_c^d g(y) , dy$. If it's entirely to the left ($g(y) le 0$), then $A = int_c^d -g(y) , dy$.
JEE Tip: Deciding whether to integrate with respect to 'x' (using $dx$) or 'y' (using $dy$) is a crucial first step. Sometimes one approach is significantly simpler than the other. Always consider both options after sketching the region.
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Case 1: Area Bounded by a Curve and the X-axis
This is the most direct application. The key is to correctly identify the limits of integration and handle cases where the curve goes below the x-axis.
Subcase 1.1: Curve entirely above or entirely below the x-axis.
If $f(x) ge 0$ for all $x in [a, b]$, Area = $int_a^b f(x) , dx$.
If $f(x) le 0$ for all $x in [a, b]$, Area = $int_a^b -f(x) , dx$.
Subcase 1.2: Curve crosses the x-axis.
If $f(x)$ changes sign within the interval $[a, b]$, say it crosses at $x=c$, then the integral $int_a^b f(x) , dx$ will give a net signed area, not the total geometric area. To find the total geometric area, you must split the integral:
Area = $int_a^b |f(x)| , dx = int_a^c |f(x)| , dx + int_c^b |f(x)| , dx$.
This translates to finding the x-intercepts and integrating section by section, taking the absolute value of each resulting integral.
Example 1: Find the area bounded by the curve $y = x^2 - 4$, the x-axis, and the lines $x=0$ and $x=3$.
Solution:
1.
Sketch the curve: $y = x^2 - 4$ is an upward-opening parabola with its vertex at $(0, -4)$ and x-intercepts at $x = pm 2$.
2.
Identify the region: We are interested in the region from $x=0$ to $x=3$.
* From $x=0$ to $x=2$, the curve $y = x^2 - 4$ is below the x-axis ($y le 0$).
* At $x=2$, the curve crosses the x-axis.
* From $x=2$ to $x=3$, the curve $y = x^2 - 4$ is above the x-axis ($y ge 0$).
3.
Set up the integral: We need to split the integral at $x=2$.
Area $A = int_0^3 |x^2 - 4| , dx = int_0^2 |x^2 - 4| , dx + int_2^3 |x^2 - 4| , dx$
$A = int_0^2 -(x^2 - 4) , dx + int_2^3 (x^2 - 4) , dx$
$A = int_0^2 (4 - x^2) , dx + int_2^3 (x^2 - 4) , dx$
4.
Evaluate the integrals:
$int (4 - x^2) , dx = 4x - frac{x^3}{3}$
$int (x^2 - 4) , dx = frac{x^3}{3} - 4x$
For the first part: $[4x - frac{x^3}{3}]_0^2 = (4(2) - frac{2^3}{3}) - (4(0) - 0) = 8 - frac{8}{3} = frac{24 - 8}{3} = frac{16}{3}$
For the second part: $[frac{x^3}{3} - 4x]_2^3 = (frac{3^3}{3} - 4(3)) - (frac{2^3}{3} - 4(2)) = (9 - 12) - (frac{8}{3} - 8) = -3 - (frac{8 - 24}{3}) = -3 - (-frac{16}{3}) = -3 + frac{16}{3} = frac{-9 + 16}{3} = frac{7}{3}$
5.
Total Area: $A = frac{16}{3} + frac{7}{3} = frac{23}{3}$ square units.
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Case 2: Area Bounded by a Curve and the Y-axis
This is analogous to Case 1, but we express $x$ as a function of $y$ ($x=g(y)$) and integrate with respect to $y$. This approach is particularly useful when vertical strips are complex to define, but horizontal strips are straightforward.
Example 2: Find the area bounded by the curve $x = y^2 - 1$, the y-axis, and the lines $y=-1$ and $y=1$.
Solution:
1.
Sketch the curve: $x = y^2 - 1$ is a parabola opening to the right, with its vertex at $(-1, 0)$ and y-intercepts at $y = pm 1$.
2.
Identify the region: We are interested in the region from $y=-1$ to $y=1$.
* In this entire interval, the curve $x = y^2 - 1$ is to the left of the y-axis ($x le 0$).
3.
Set up the integral: Since the curve is to the left of the y-axis, $x le 0$, we integrate $-x , dy$.
Area $A = int_{-1}^1 -(y^2 - 1) , dy = int_{-1}^1 (1 - y^2) , dy$
4.
Evaluate the integral:
$int (1 - y^2) , dy = y - frac{y^3}{3}$
$A = [y - frac{y^3}{3}]_{-1}^1 = (1 - frac{1^3}{3}) - (-1 - frac{(-1)^3}{3})$
$A = (1 - frac{1}{3}) - (-1 - (-frac{1}{3})) = (frac{2}{3}) - (-1 + frac{1}{3}) = frac{2}{3} - (-frac{2}{3}) = frac{2}{3} + frac{2}{3} = frac{4}{3}$ square units.
Note: This parabola is symmetric about the x-axis. We could have calculated $2 int_0^1 (1 - y^2) , dy$ to simplify calculations.
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Case 3: Area Bounded by Two Curves
This is a very common scenario in JEE. The principle is to find the area between the "top" curve and the "bottom" curve (for $dx$ integration) or the "right" curve and the "left" curve (for $dy$ integration).
If $y_1 = f(x)$ and $y_2 = g(x)$ are two curves, and $f(x) ge g(x)$ in the interval $[a, b]$, then the area between them is:
$A = int_a^b (f(x) - g(x)) , dx$
Here, $f(x)$ is the "upper" curve and $g(x)$ is the "lower" curve.
Similarly, if $x_1 = f(y)$ and $x_2 = g(y)$ are two curves, and $f(y) ge g(y)$ in the interval $[c, d]$, then the area between them is:
$A = int_c^d (f(y) - g(y)) , dy$
Here, $f(y)$ is the "right" curve and $g(y)$ is the "left" curve.
Caution: Always find the points of intersection of the curves first, as these often define the limits of integration. A clear sketch is indispensable. If curves cross each other within the region, you'll need to split the integral, similar to Case 1.
Example 3: Find the area bounded by the curves $y = x^2$ and $y = x$.
Solution:
1.
Sketch the curves: $y = x^2$ is an upward-opening parabola through the origin. $y=x$ is a straight line through the origin with a slope of 1.
2.
Find points of intersection: Set $x^2 = x$.
$x^2 - x = 0 Rightarrow x(x-1) = 0$.
So, $x=0$ and $x=1$.
The corresponding y-values are $y=0^2=0$ and $y=1^2=1$.
Intersection points are $(0,0)$ and $(1,1)$.
3.
Identify upper and lower curves: In the interval $[0,1]$:
Test a point, e.g., $x=0.5$.
For $y=x$, $y=0.5$.
For $y=x^2$, $y=(0.5)^2=0.25$.
Since $0.5 > 0.25$, the line $y=x$ is the upper curve, and the parabola $y=x^2$ is the lower curve.
4.
Set up the integral:
Area $A = int_0^1 (x - x^2) , dx$
5.
Evaluate the integral:
$int (x - x^2) , dx = frac{x^2}{2} - frac{x^3}{3}$
$A = [frac{x^2}{2} - frac{x^3}{3}]_0^1 = (frac{1^2}{2} - frac{1^3}{3}) - (0 - 0) = frac{1}{2} - frac{1}{3} = frac{3-2}{6} = frac{1}{6}$ square units.
Example 4: Find the area bounded by $y^2 = 4ax$ and $x^2 = 4by$. (A classic JEE problem!)
Solution:
1.
Sketch the curves: $y^2 = 4ax$ is a parabola opening right (assuming $a>0$). $x^2 = 4by$ is a parabola opening upward (assuming $b>0$).
2.
Find points of intersection:
From $y^2 = 4ax$, we have $y = sqrt{4ax} = 2sqrt{ax}$ (considering the upper half for simplicity, we'll multiply by 2 if needed, but here we can substitute directly).
Substitute $y = frac{x^2}{4b}$ (from $x^2 = 4by$) into $y^2 = 4ax$:
$(frac{x^2}{4b})^2 = 4ax$
$frac{x^4}{16b^2} = 4ax$
$x^4 = 64ab^2x$
$x^4 - 64ab^2x = 0$
$x(x^3 - 64ab^2) = 0$
So, $x=0$ is one intersection point.
For the other, $x^3 = 64ab^2 Rightarrow x = sqrt[3]{64ab^2} = 4a^{1/3}b^{2/3}$.
Let $x_0 = 4a^{1/3}b^{2/3}$.
When $x=0$, $y=0$.
When $x=x_0$, $y = frac{x_0^2}{4b} = frac{(4a^{1/3}b^{2/3})^2}{4b} = frac{16a^{2/3}b^{4/3}}{4b} = 4a^{2/3}b^{1/3}$.
So, the intersection points are $(0,0)$ and $(4a^{1/3}b^{2/3}, 4a^{2/3}b^{1/3})$.
3.
Identify upper and lower curves:
For the region between $x=0$ and $x=x_0$, we need to compare $y_1 = sqrt{4ax} = 2sqrt{ax}$ and $y_2 = frac{x^2}{4b}$.
Let's check a point, say $x=a$ (assuming $x_0 > a$).
$y_1 = 2a$.
$y_2 = frac{a^2}{4b}$.
It's not immediately obvious which is greater without knowing $a$ and $b$. A better way is to consider the power of $x$. For small $x$, $sqrt{x}$ grows faster than $x^2$. So, for $x in (0, x_0)$, $y^2=4ax$ (or $y=sqrt{4ax}$) is the upper curve. Let's verify: for $x in (0, x_0)$, $x^2 < 4ax$ and $sqrt{4ax} > frac{x^2}{4b}$.
Alternatively, for the intersection $x = 4a^{1/3}b^{2/3}$, $y^2 = 4ax$ gives $y = 2sqrt{ax}$, and $x^2 = 4by$ gives $y = frac{x^2}{4b}$.
At any $x$ between $0$ and $x_0$, $2sqrt{ax} ge frac{x^2}{4b}$.
So, the upper curve is $y_1 = 2sqrt{ax}$ and the lower curve is $y_2 = frac{x^2}{4b}$.
4.
Set up the integral:
Area $A = int_0^{x_0} (2sqrt{ax} - frac{x^2}{4b}) , dx$
$A = int_0^{4a^{1/3}b^{2/3}} (2sqrt{a}x^{1/2} - frac{1}{4b}x^2) , dx$
5.
Evaluate the integral:
$int (2sqrt{a}x^{1/2} - frac{1}{4b}x^2) , dx = 2sqrt{a} frac{x^{3/2}}{3/2} - frac{1}{4b} frac{x^3}{3} = frac{4sqrt{a}}{3} x^{3/2} - frac{x^3}{12b}$
Now, substitute the limits: $x_0 = 4a^{1/3}b^{2/3}$.
$A = [frac{4sqrt{a}}{3} x^{3/2} - frac{x^3}{12b}]_0^{x_0}$
$A = frac{4sqrt{a}}{3} (x_0)^{3/2} - frac{(x_0)^3}{12b}$
$(x_0)^{3/2} = (4a^{1/3}b^{2/3})^{3/2} = 4^{3/2} (a^{1/3})^{3/2} (b^{2/3})^{3/2} = 8 a^{1/2} b^{1} = 8sqrt{a}b$
$(x_0)^3 = (4a^{1/3}b^{2/3})^3 = 4^3 (a^{1/3})^3 (b^{2/3})^3 = 64ab^2$
$A = frac{4sqrt{a}}{3} (8sqrt{a}b) - frac{64ab^2}{12b}$
$A = frac{32a b}{3} - frac{16ab}{3}$
$A = frac{16ab}{3}$ square units.
JEE Standard Result: The area enclosed by the parabolas $y^2 = 4ax$ and $x^2 = 4by$ is always $frac{16ab}{3}$ square units. Memorizing this can save time in JEE Mains, but knowing the derivation is crucial for Advanced.
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Areas of Standard Regions (JEE Favorites)
These are frequently encountered shapes. Understanding their derivations solidifies your conceptual base.
1.
Area of a Circle: $x^2 + y^2 = a^2$
The total area is 4 times the area in the first quadrant.
In the first quadrant, $y = sqrt{a^2 - x^2}$ from $x=0$ to $x=a$.
Area $= 4 int_0^a sqrt{a^2 - x^2} , dx$
Using the standard integral formula $int sqrt{a^2 - x^2} , dx = frac{x}{2}sqrt{a^2-x^2} + frac{a^2}{2}sin^{-1}left(frac{x}{a}
ight)$:
Area $= 4 left[ frac{x}{2}sqrt{a^2-x^2} + frac{a^2}{2}sin^{-1}left(frac{x}{a}
ight)
ight]_0^a$
Area $= 4 left[ left(frac{a}{2}sqrt{a^2-a^2} + frac{a^2}{2}sin^{-1}left(frac{a}{a}
ight)
ight) - left(0 + frac{a^2}{2}sin^{-1}(0)
ight)
ight]$
Area $= 4 left[ left(0 + frac{a^2}{2}cdotfrac{pi}{2}
ight) - (0)
ight] = 4 cdot frac{pi a^2}{4} = pi a^2$.
2.
Area of an Ellipse: $frac{x^2}{a^2} + frac{y^2}{b^2} = 1$
The total area is 4 times the area in the first quadrant.
In the first quadrant, $y = frac{b}{a}sqrt{a^2 - x^2}$ from $x=0$ to $x=a$.
Area $= 4 int_0^a frac{b}{a}sqrt{a^2 - x^2} , dx = frac{4b}{a} int_0^a sqrt{a^2 - x^2} , dx$
Using the same integral as for the circle:
Area $= frac{4b}{a} left[ frac{x}{2}sqrt{a^2-x^2} + frac{a^2}{2}sin^{-1}left(frac{x}{a}
ight)
ight]_0^a$
Area $= frac{4b}{a} left[ frac{a^2}{2}cdotfrac{pi}{2}
ight] = frac{4b}{a} cdot frac{pi a^2}{4} = pi ab$.
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Working with Absolute Values
When a function involves absolute values, like $y = |f(x)|$ or $y = |f(x) - g(x)|$, it's crucial to break down the integral at the points where the expression inside the absolute value changes sign. This is because $|X|$ is defined piecewise: $X$ if $X ge 0$, and $-X$ if $X < 0$.
Example 5: Find the area bounded by $y = |cos x|$, the x-axis, from $x=0$ to $x=2pi$.
Solution:
1.
Sketch the curve: The graph of $y = |cos x|$ is the graph of $cos x$ but with the negative parts flipped above the x-axis.
2.
Identify where $cos x$ changes sign:
* $cos x ge 0$ for $x in [0, pi/2]$ and $x in [3pi/2, 2pi]$.
* $cos x < 0$ for $x in (pi/2, 3pi/2)$.
3.
Set up the integral:
Area $A = int_0^{2pi} |cos x| , dx$
$A = int_0^{pi/2} cos x , dx + int_{pi/2}^{3pi/2} (-cos x) , dx + int_{3pi/2}^{2pi} cos x , dx$
4.
Evaluate the integrals:
$int cos x , dx = sin x$
$int (-cos x) , dx = -sin x$
Part 1: $[sin x]_0^{pi/2} = sin(pi/2) - sin(0) = 1 - 0 = 1$.
Part 2: $[-sin x]_{pi/2}^{3pi/2} = (-sin(3pi/2)) - (-sin(pi/2)) = (-(-1)) - (-1) = 1 + 1 = 2$.
Part 3: $[sin x]_{3pi/2}^{2pi} = sin(2pi) - sin(3pi/2) = 0 - (-1) = 1$.
5.
Total Area: $A = 1 + 2 + 1 = 4$ square units.
JEE Tip: For functions like $|sin x|$ or $|cos x|$, recognize their periodic nature and symmetry. For example, $int_0^{2pi} |sin x| , dx = 4 int_0^{pi/2} sin x , dx = 4[-cos x]_0^{pi/2} = 4(0 - (-1)) = 4$.
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Important Considerations for JEE Advanced:
1.
Sketching is your Superpower: Always, always draw a clear sketch of the region. It helps you:
* Identify the limits of integration.
* Determine which curve is 'above/below' or 'right/left'.
* Spot any symmetry that can simplify calculations.
* Catch errors if your integral result doesn't make sense with the sketch.
2.
Choice of Integration Variable (dx vs dy):
Sometimes, integrating with respect to $y$ (horizontal strips) can transform a complex integral into a much simpler one.
Example: Area of a region bounded by $x=y^2$ and $x=4$.
If you integrate with respect to $x$: $y=pmsqrt{x}$. You'd need $int_0^4 (sqrt{x} - (-sqrt{x})) , dx = int_0^4 2sqrt{x} , dx$.
If you integrate with respect to $y$: $x=4$ is the right curve, $x=y^2$ is the left curve. Limits for $y$ are from $y=-2$ to $y=2$.
Area $= int_{-2}^2 (4 - y^2) , dy$. This is often easier.
3.
Symmetry: Exploit symmetry whenever possible. If a region is symmetric about the x-axis, y-axis, or origin, you can calculate the area of one part and multiply by an appropriate factor. For example, for a circle or an ellipse, we calculated 4 times the area in the first quadrant.
4.
Standard Integral Formulas: Be extremely proficient with integrals involving $sqrt{a^2-x^2}$, $sqrt{x^2-a^2}$, $sqrt{x^2+a^2}$. These appear very frequently in area problems, especially involving circles, ellipses, and hyperbolas (though hyperbolas are less common for bounded regions).
5.
Area by Subtraction: Sometimes, it's easier to find the area of a larger, simpler region and subtract the area of a smaller, unwanted region. This is particularly useful when boundaries are complex.
By mastering these techniques and practicing diligently with a variety of problems, you'll be well-prepared to tackle any area-related question in your JEE exams. Keep practicing, keep visualizing, and keep integrating!
