Area under Simple Curves

Geometrical Interpretation of Definite Integrals

In the previous chapter, we learned that the definite integral $\int_a^b f(x) \, dx$ represents the limit of a sum. Geometrically, if $f(x)$ is a continuous and non-negative function on the interval $[a, b]$, then this definite integral gives the exact area of the region bounded by the curve $y = f(x)$, the x-axis, and the vertical lines $x = a$ and $x = b$.

To understand this visually, imagine dividing the region into a very large number of thin vertical strips. For a strip at position $x$:

• its height is $y = f(x)$,
• its width is $dx$,
• so its small area is $dA = y\,dx$.

Adding all such strips from $x=a$ to $x=b$ gives the total area: $$\text{Area} = \int_a^b y\,dx = \int_a^b f(x)\,dx$$

This interpretation is valid directly when the curve lies on or above the x-axis throughout the interval. If the curve goes below the axis, then the definite integral gives signed area, not total geometrical area.

Area Bounded by the y-axis

Sometimes it is more convenient to integrate with respect to $y$ instead of $x$.

If the curve is given in the form $x = g(y)$, and the required region is bounded by:

• the curve $x = g(y)$,
• the y-axis,
• the horizontal lines $y = c$ and $y = d$,

then we use horizontal strips.

For a strip at height $y$:

• its length is $x = g(y)$,
• its thickness is $dy$,
• so its small area is $dA = x\,dy$.

Hence the total area is: $$\text{Area} = \int_c^d x\,dy = \int_c^d g(y)\,dy$$

This method is especially useful when the curve is naturally expressed as $x$ in terms of $y$, or when using vertical strips would be cumbersome.

Areas Below the Coordinate Axes

If the curve $y = f(x)$ lies below the x-axis on an interval $[a,b]$, then $f(x) < 0$ there. In that case, $$\int_a^b f(x)\,dx$$ is negative, because the definite integral measures signed area.

But geometrical area can never be negative. Therefore, if the whole curve lies below the x-axis, the required area is: $$\text{Area} = \left|\int_a^b f(x)\,dx\right|$$

When the curve crosses the x-axis: If the curve intersects the x-axis at some point $x=c$ inside the interval, then the positive and negative parts do not represent cancellation in geometry. So we must split the integral at the point of intersection and add the magnitudes of the separate parts.

If $f(x)<0$ on $[a,c]$ and $f(x)>0$ on $[c,b]$, then: $$\text{Total Area} = \left|\int_a^c f(x)\,dx\right| + \int_c^b f(x)\,dx$$

In general, always:

1. find where the curve cuts the axis,
2. split the interval there,
3. evaluate each part separately,
4. add positive areas only.

Example 1: Area of a Simple Parabola

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

Solution: Step 1: Identify the curve and the interval. The curve is $y=x^2$, and the vertical boundaries are $x=1$ and $x=2$. Since $x^2>0$ for all $x\neq 0$, the curve lies above the x-axis on $[1,2]$.

Step 2: Write the area integral. $$\text{Area} = \int_1^2 x^2\,dx$$

Step 3: Integrate. $$\int x^2\,dx = \frac{x^3}{3}$$ So, $$\text{Area} = \left[\frac{x^3}{3}\right]_1^2$$

Step 4: Apply the limits. $$= \frac{2^3}{3} - \frac{1^3}{3} = \frac{8}{3} - \frac{1}{3} = \frac{7}{3}$$

Answer: $\frac{7}{3}$ square units.

Example 2: Area bounded by the y-axis

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

Solution: Step 1: Notice the orientation of the boundaries. The region is bounded by the y-axis and two horizontal lines, so integrating with respect to $y$ is the natural choice.

Step 2: The curve is already in the form $x=g(y)$ with $$x = 2y$$ Thus the required area is $$\text{Area} = \int_1^3 x\,dy = \int_1^3 2y\,dy$$

Step 3: Integrate. $$\int 2y\,dy = y^2$$ Therefore, $$\text{Area} = \left[y^2\right]_1^3$$

Step 4: Evaluate. $$= 3^2 - 1^2 = 9 - 1 = 8$$

Answer: $8$ square units.

Example 3: Area of a Circle using Integration

Find the area of the region bounded by the circle $x^2 + y^2 = a^2$.

Solution: Step 1: Use symmetry. The circle is symmetric about both coordinate axes, so it is enough to find the area in the first quadrant and multiply by 4.

Step 2: Express $y$ in terms of $x$. From $$x^2 + y^2 = a^2$$ we get $$y = \sqrt{a^2 - x^2}$$ for the upper semicircle. In the first quadrant, $x$ varies from $0$ to $a$.

Step 3: Write the integral for one quadrant. $$\text{Area of first quadrant} = \int_0^a \sqrt{a^2 - x^2}\,dx$$ Hence total area is $$\text{Total Area} = 4\int_0^a \sqrt{a^2-x^2}\,dx$$

Step 4: Use the standard result: $$\int \sqrt{a^2-x^2}\,dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)$$ So, $$\text{Total Area} = 4\left[\frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)\right]_0^a$$

Step 5: Substitute the limits. At $x=a$: $$\frac{a}{2}\sqrt{a^2-a^2} + \frac{a^2}{2}\sin^{-1}(1) = 0 + \frac{a^2}{2}\cdot \frac{\pi}{2} = \frac{\pi a^2}{4}$$ At $x=0$: $$0 + \frac{a^2}{2}\sin^{-1}(0)=0$$

Step 6: Multiply by 4. $$\text{Total Area} = 4\cdot \frac{\pi a^2}{4} = \pi a^2$$

Answer: $\pi a^2$ square units.

Example 4: Area of an Ellipse

Find the area of the region bounded by the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

Solution: Step 1: Use symmetry. The ellipse is symmetric about both axes, so total area equals 4 times the area in the first quadrant.

Step 2: Express $y$ in terms of $x$. From $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ we get $$\frac{y^2}{b^2} = 1 - \frac{x^2}{a^2} = \frac{a^2-x^2}{a^2}$$ So, $$y = \frac{b}{a}\sqrt{a^2-x^2}$$ for the first quadrant.

Step 3: Write the area integral. $$\text{Total Area} = 4\int_0^a \frac{b}{a}\sqrt{a^2-x^2}\,dx = \frac{4b}{a}\int_0^a \sqrt{a^2-x^2}\,dx$$

Step 4: Use the standard result from the circle problem. $$\int_0^a \sqrt{a^2-x^2}\,dx = \frac{\pi a^2}{4}$$ Hence, $$\text{Total Area} = \frac{4b}{a}\cdot \frac{\pi a^2}{4} = \pi ab$$

Answer: $\pi ab$ square units.

Example 5: Area Below and Above the x-axis

Find the area bounded by the curve $y = x^3$, the x-axis, and the ordinates $x = -2$ and $x = 1$.

Solution: Step 1: Determine where the curve crosses the x-axis. For $y=x^3$, the curve crosses the x-axis at $$x^3=0 \implies x=0$$ So the interval must be split at $x=0$.

Step 2: Check the sign of the function.

• On $[-2,0]$, $x^3<0$, so the curve lies below the x-axis.
• On $[0,1]$, $x^3>0$, so the curve lies above the x-axis.

Step 3: Write the area as sum of magnitudes. $$\text{Total Area} = \left|\int_{-2}^0 x^3\,dx\right| + \int_0^1 x^3\,dx$$

Step 4: Evaluate each integral. $$\int x^3\,dx = \frac{x^4}{4}$$ So, $$\int_{-2}^0 x^3\,dx = \left[\frac{x^4}{4}\right]_{-2}^0 = 0 - \frac{16}{4} = -4$$ Therefore its area contribution is $|-4|=4$.

Also, $$\int_0^1 x^3\,dx = \left[\frac{x^4}{4}\right]_0^1 = \frac14$$

Step 5: Add the two positive areas. $$\text{Total Area} = 4 + \frac14 = \frac{17}{4}$$

Answer: $\frac{17}{4}$ square units.

Example 6: Area Bounded by a Trigonometric Curve

Find the area of the region bounded by the curve $y = \cos x$ between $x = 0$ and $x = 2\pi$.

Solution: Step 1: Identify where the curve changes sign. The graph of $y=\cos x$ crosses the x-axis at $$x=\frac{\pi}{2},\ \frac{3\pi}{2}$$ Thus the interval $[0,2\pi]$ must be split into three parts.

Step 2: Determine the sign in each part.

• On $\left[0,\frac\pi2\right]$, $\cos x>0$.
• On $\left[\frac\pi2,\frac{3\pi}2\right]$, $\cos x<0$.
• On $\left[\frac{3\pi}2,2\pi\right]$, $\cos x>0$.

Step 3: Write the total area. $$\text{Total Area} = \int_0^{\pi/2} \cos x\,dx + \left|\int_{\pi/2}^{3\pi/2} \cos x\,dx\right| + \int_{3\pi/2}^{2\pi} \cos x\,dx$$

Step 4: Evaluate each part. Since $$\int \cos x\,dx = \sin x,$$ we get $$\int_0^{\pi/2} \cos x\,dx = [\sin x]_0^{\pi/2}=1$$ $$\int_{\pi/2}^{3\pi/2} \cos x\,dx = [\sin x]_{\pi/2}^{3\pi/2}=(-1)-1=-2$$ So its area is $|-2|=2$. $$\int_{3\pi/2}^{2\pi} \cos x\,dx = [\sin x]_{3\pi/2}^{2\pi}=0-(-1)=1$$

Step 5: Add the areas. $$\text{Total Area} = 1+2+1=4$$

Answer: $4$ square units.

Example 7: Area Involving an Absolute Value Function

Find the area of the region bounded by the curve $y = |x+1|$, the lines $x = -3$ and $x = 1$, and the x-axis.

Solution: Step 1: Write the function in piecewise form. $$|x+1| = \begin{cases} -(x+1), & x<-1 \\ x+1, & x\ge -1 \end{cases}$$ The expression inside the modulus becomes zero at $x=-1$, so we split the interval there.

Step 2: Write the area as two integrals. $$\text{Area} = \int_{-3}^{-1} -(x+1)\,dx + \int_{-1}^{1} (x+1)\,dx$$ Since the absolute value graph is never below the x-axis, these integrals already give positive area.

Step 3: Evaluate the first integral. $$\int_{-3}^{-1} ( -x-1)\,dx = \left[-\frac{x^2}{2}-x\right]_{-3}^{-1}$$ At $x=-1$: $$-\frac{1}{2}+1 = \frac12$$ At $x=-3$: $$-\frac{9}{2}+3 = -\frac32$$ So, $$\frac12 - \left(-\frac32\right)=2$$

Step 4: Evaluate the second integral. $$\int_{-1}^{1} (x+1)\,dx = \left[\frac{x^2}{2}+x\right]_{-1}^{1}$$ At $x=1$: $$\frac12+1 = \frac32$$ At $x=-1$: $$\frac12-1 = -\frac12$$ So, $$\frac32 - \left(-\frac12\right)=2$$

Step 5: Add the two parts. $$\text{Total Area} = 2+2=4$$

Answer: $4$ square units.

Example 8: Area of an Exponential Curve

Find the area bounded by the curve $y = e^x$, the x-axis, and the ordinates $x = 0$ and $x = 2$.

Solution: Step 1: Observe that $e^x>0$ for all real $x$, so the curve remains above the x-axis on $[0,2]$.

Step 2: Write the area integral. $$\text{Area} = \int_0^2 e^x\,dx$$

Step 3: Integrate. $$\int e^x\,dx = e^x$$ Thus, $$\text{Area} = [e^x]_0^2$$

Step 4: Evaluate the limits. $$= e^2 - e^0 = e^2 - 1$$

Answer: $e^2 - 1$ square units.

Example 9: Area involving Fractional Powers

Find the area bounded by the curve $y = 2\sqrt{x}$, the x-axis, and the lines $x = 1$ and $x = 4$.

Solution: Step 1: Since $2\sqrt{x}>0$ for $x>0$, the required area is directly $$\text{Area} = \int_1^4 2\sqrt{x}\,dx = 2\int_1^4 x^{1/2}\,dx$$

Step 2: Use the power rule. $$\int x^{1/2}\,dx = \frac{x^{3/2}}{3/2} = \frac{2}{3}x^{3/2}$$ Hence, $$\text{Area} = 2\left[\frac{2}{3}x^{3/2}\right]_1^4 = \frac{4}{3}[x^{3/2}]_1^4$$

Step 3: Evaluate the powers. $$4^{3/2} = (\sqrt4)^3 = 2^3 = 8, \qquad 1^{3/2}=1$$ So, $$\text{Area} = \frac43(8-1)=\frac43\cdot 7 = \frac{28}{3}$$

Answer: $\frac{28}{3}$ square units.

Example 10: Area bounded by a Logarithmic Curve

Find the area bounded by the curve $y = \ln x$, the x-axis, and the lines $x = 1$ and $x = e$.

Solution: Step 1: Check the sign of $\ln x$ on the interval. For $1\le x\le e$, we have $\ln x \ge 0$, so the curve lies on or above the x-axis.

Step 2: Write the integral. $$\text{Area} = \int_1^e \ln x\,dx$$

Step 3: Integrate by parts. Take $$u=\ln x, \qquad dv=dx$$ Then $$du=\frac{1}{x}dx, \qquad v=x$$ So, $$\int \ln x\,dx = x\ln x - \int x\cdot \frac1x\,dx = x\ln x - \int 1\,dx = x\ln x - x$$

Step 4: Apply the limits. $$\text{Area} = [x\ln x - x]_1^e$$ At $x=e$: $$e\ln e - e = e-e=0$$ At $x=1$: $$1\ln 1 -1 = 0-1=-1$$ Therefore, $$\text{Area} = 0 - (-1) = 1$$

Answer: $1$ square unit.

Example 11: Area of an Intercept Region

Find the area of the region bounded by $y = x^2 - 4$ and the x-axis.

Solution: Step 1: Find where the curve cuts the x-axis. Set $y=0$: $$x^2-4=0 \implies x=\pm 2$$ So the bounded region lies between $x=-2$ and $x=2$.

Step 2: Check whether the curve is above or below the x-axis in this interval. At $x=0$, $$y=0^2-4=-4<0$$ So the curve lies below the x-axis on $[-2,2]$.

Step 3: Write the area using modulus. $$\text{Area} = \left|\int_{-2}^{2}(x^2-4)\,dx\right|$$

Step 4: Evaluate the integral. $$\int (x^2-4)\,dx = \frac{x^3}{3} - 4x$$ Hence, $$\int_{-2}^{2}(x^2-4)\,dx = \left[\frac{x^3}{3}-4x\right]_{-2}^{2}$$ At $x=2$: $$\frac{8}{3}-8 = -\frac{16}{3}$$ At $x=-2$: $$-\frac{8}{3}+8 = \frac{16}{3}$$ So, $$-\frac{16}{3} - \frac{16}{3} = -\frac{32}{3}$$

Step 5: Take the absolute value. $$\text{Area} = \left|-\frac{32}{3}\right| = \frac{32}{3}$$

Answer: $\frac{32}{3}$ square units.