Understanding Definite Integration
Properties of Definite Integration
| 1 . | Order of the limit of integration | \(\int_a^b f(x) dx = – \int_b^a f(x) dx\) |
| 2. | Addition/Subtraction | \(\int_a^b \left(f(x) \pm g(x) \right) dx = \int_a^b f(x) dx \pm \int_a^b g(x) dx\) |
| 3. |
Additive | \(\int_a^b f(x) dx + \int_b^c f(x) dx = \int_a^c f(x) dx \) |
| 4. | Constant Multiples | \(\int_a^b k f(x) dx = k \int_a^b f(x) dx \) |
| 5. | Zero width Interval | \(\int_a^a f(x) dx = 0 \) |
Calculation of definite Integrations
The Fundamental Theorem of Calculus-part 1:
If \(f\) is continuous on \([a, b]\), then \(F(x) = \int_a^x f(t) dt\) is continuous on \([a, b]\) and differentiable on \( (a, b) \), and the derivative of \(F(x) \) is \(f(x)\) given as \[ F'(x) = \frac{d}{dx}\int_a^x f(t) dt = f(x). \]
Important Points to Remember
Following points need to remember for proper application of The Fundamental Theorem of Calculus-part 1.
- The lower limit of the definite integration should be constant and the upper limit is either the variable or a function of the concern variable,
- If the upper limit is a function of the concern variable, say \(g(x)\), then \[ F'(x) = \frac{d}{dx}\int_a^{g(x)} f(t) dt = f(g(x))\frac{d g(x)}{dx}.\]
Examples related with The Fundamental Theorem of Calculus- Part 1
Example-1
Find \(dy/dx\) from each of the problem
- \( y = \int_3^x \sin^3(t) dt \implies \frac{dy}{dx}= \sin^3(x)\)
- \( y = \int_3^{x^2} \sin^3(t) dt \implies \frac{dy}{dx}= \sin^3(x) \frac{d x^2}{dx} = 2 x \sin^3(x) \)
(Carefully note the difference between (a) and (b), just changing the upper limit from \(x\) to \(x^2\) how the calculation change) - \( y = \int_{e^x}^3 \sin^3(t) dt \). Clearly, the given problem is not in the standard form to apply the Fundamental Theorem. So, by using the properties of the definite integration, we have \[ y = -\int_3^{e^x} \sin^3(t) dt \implies \frac{dy}{dx} = – \sin^3(e^x) \frac{d}{dx}e^x = – \;e^x \sin^3(e^x). \]
Example-2
Find \(dy/dx\) from
\( y = \int_{x^2}^{x^3} \sqrt{3t^2+1} dt \).
To use fundamental theorem, in this case, we have to break the integration into two parts by taking a constant in the middle. However, we can do this directly as follow: \[\begin{aligned} \frac{dy}{dx} &= \frac{d}{dx} \int_{x^2}^{x^3} \sqrt{3t^2+1} dt \\ &= \sqrt{3(x^3)^2+1} \;\frac{d}{dx} (x^3)\; -\; \sqrt{3(x^2)^2+1} \;\frac{d}{dx} (x^2)\\ &= 3 x^2 \sqrt{3 x^6 +1} \; – \quad 2 x \sqrt{3 x^4+1}. \end{aligned}\]
Example-3
Find \(dy/dx\) from
\( y = \int_{2 \tan(x)}^{0} \frac{1}{4+t^2} dt \).
\[\begin{aligned} \frac{dy}{dx} &= \frac{d}{dx} \int_{2 \tan(x)}^{0} \frac{1}{4+t^2} dt \\ &= – \frac{d}{dx} \int_{0}^{2 \tan(x)} \frac{1}{4+t^2} dt \\ &= – \frac{1}{4 + (2 \tan(x))^2} \;\frac{d (2 \tan(x))}{dx} \\ &= – \frac{2 \sec^2(x)}{4 (1 + \tan^2(x))} = – \frac{2 \sec^2(x)}{4 \sec^2(x)}= – \frac{1}{2}. \end{aligned}\]
Example-4
Find \(dy/dx\) from
\(y = \int_{3^{x}}^{1} \ln(t) dt \).
\[\begin{aligned} \frac{dy}{dx} &= \frac{d}{dx} \int_{3^{x}}^{1} \ln(t) dt \\ &= – \frac{d}{dx} \int_{1}^{3^x} \ln(t) dt \\ &= – \ln(3^x)\;\frac{d (3^x)}{dx} \\ &= -\left( x \ln(3) \right) \; \left(3^x \ln(3) \right)= – x\; 3^x \; \left(\ln(3)\right)^2. \end{aligned}\]
The Fundamental Theorem of Calculus-part 2:
If \(f\) is continuous on \([a, b]\), and \(F(x) \) is any antiderivative of \(f(x)\), then \[\int_a^b f(x) dx = \bigg(F(x)\bigg)_a^b= F(b)\; -\; F(a). \]
Examples Based on the Fundamental Theorem of Calculus-part 2
Example
Evaluate the following definite integralsNote: For all the example below, you need Antiderivatives
\[\int_0^4 x^2 dx = \bigg( \frac{x^3}{3} \bigg)_0^4 = \frac{4^3}{3} – \frac{0^3}{3} = \frac{64}{3}. \]
\[ \int_0^{\pi} \sin(2x) dx = \bigg( – \frac{\cos(2x)}{2} \bigg)_0^{\pi} = \frac{\cos(2\pi)}{2} – \frac{\cos(0)}{2} =\frac{1}{2} – \frac{1}{2}=0. \quad \] \[\cos(2\pi)= 1, \cos(0)=1\]
\[ \int_0^{1} \frac{3}{1+ \theta^2} d\theta = \bigg( 3 \tan^{-1}(\theta) \bigg)_0^{1} = 3 \bigg(\tan^{-1}(1)- \tan^{-1}(0) \bigg) =3 \bigg(\frac{\pi}{4}- 0 \bigg) = \frac{3 \pi}{4}. \quad \] \[\tan^{-1}(1)= \frac{\pi}{4}, \tan^{-1}(0)=0\]
\[ \begin{aligned} \int_{-4}^{2}\left( 6 \; – \frac{t}{3} \right)dt = \bigg( 6t – \frac{t^2}{6} \bigg)_{-4}^{2}&= \bigg( ( 6 \times 2) – \frac{2^2}{6} \bigg)- \bigg( ( 6 \times (-4)) – \frac{(-4)^2}{6} \bigg)\\ &= \bigg( 12\; – \frac{2}{3} \bigg)- \bigg( – 24 – \frac{8}{3} \bigg) \\ & = 12\; – \frac{2}{3}+ 24 + \frac{8}{3} = 36 + 2= 38. \end{aligned} \]
\[ \begin{aligned} \int_{-1}^{1}\left( 3 x^2 – 2x + 4 \right)dx = \bigg( \frac{3 x^3}{3} – \frac{2 x^2}{2} + 4 x\bigg)_{-1}^{1}&= \bigg( x^3 – x^2 + 4 x \bigg)_{-1}^{1}\\ &= \bigg( 1^3 – 1^2 + 4\bigg)- \bigg( (-1)^3 – (-1)^2 + 4. (-1) \bigg) \\ & = 4\; – (-6) = 10. \end{aligned} \]
\[ \int_{1}^{64} x^{-\frac{7}{6}} dx = \bigg( \frac{x^{-\frac{7}{6}+1}}{-\frac{7}{6}+1} \bigg)_{1}^{64} = \bigg( \frac{(64)^{-\frac{1}{6}}}{-\frac{1}{6}} – \frac{(1)^{-\frac{1}{6}}}{-\frac{1}{6}} \bigg) = \bigg( \frac{(2^6)^{-\frac{1}{6}}}{-\frac{1}{6}}\; – \;\frac{(1)^{-\frac{1}{6}}}{-\frac{1}{6}} \bigg) = – 3+ 6 = 3. \]
\[ \begin{aligned}\int_0^{2\pi} \bigg(1 + \cos\left(\frac{x}{2}\right)\bigg) dx &= \bigg( x + 2 \sin\left(\frac{x}{2}\right)\bigg)_{0}^{2 \pi} \\&= \bigg( 2 \pi + 2 \sin\left(\frac{2 \pi}{2}\right)\bigg) – \bigg( 0 + 2 \sin\left(0\right)\bigg) = 2 \pi.\end{aligned} \]
\[ \int_{0}^{\pi/6} \sec(2 u) \tan( 2 u) du = \bigg( \frac{\sec(2 u)}{2} \bigg)_0^{\pi/6}= \frac{1}{2} \bigg( \sec\left(\frac{\pi}{3}\right)- \sec(0) \bigg) = \frac{1}{2} \bigg( 2 – 1\bigg) = \frac{1}{2}. \]
\[ \begin{aligned}\int_{-\pi/3}^{\pi/3} \frac{1+\cos(2t)}{2} dt &= \frac{1}{2}\bigg( t + \frac{\sin(2t)}{2} \bigg)_{-\pi/3}^{\pi/3}\\&= \frac{1}{2} \bigg( \left(\frac{\pi}{3}+ \frac{\sin(\frac{2\pi}{3})}{2} \right)-\left(-\frac{\pi}{3}+ \frac{\sin(\frac{-2\pi}{3})}{2} \right)\bigg) \\&= \frac{1}{2} \bigg( \frac{2\pi}{3} + \frac{\sqrt{3}}{2} – (-\frac{\sqrt{3}}{2})\bigg) = \frac{2\pi}{3}+\frac{\sqrt{3}}{2}. \end{aligned}\]