Computation and Properties of Definite Integrals
After learning about Riemann sums and the Fundamental Theorem of Calculus, in this article we will specifically discuss the material on Computation and Properties of Definite Integrals. Actually, the method of computing definite integrals uses the Fundamental Theorem of Calculus II (FTC II) which applies generally to all types of functions \( f(x) \).
In performing definite integral computation, it's actually easy because we just need to substitute the upper and lower bounds into the integrated function. So, the emphasis remains on indefinite integrals. Therefore, it's good for us to learn and master integration techniques like algebraic function integration, trigonometric function integration, and several integration techniques: algebraic substitution, integration by parts, trigonometric substitution, and partial fractions.
After being able to compute indefinite integrals, then we substitute the bounds. Please also read the material "What's the Difference Between Definite and Indefinite Integrals".
Computation of Definite Integrals
To compute definite integrals, we use the Fundamental Theorem of Calculus II:
If \( f \) is continuous on \( [a, b] \) and \( F \) is an antiderivative of \( f \) on \( [a, b] \), then:
Compute: \(\int_0^2 (3x^2 - 4x + 1) dx\)
Solution:
First, find the antiderivative:
\(\int (3x^2 - 4x + 1) dx = x^3 - 2x^2 + x + C\)
Now apply FTC II:
\(\int_0^2 (3x^2 - 4x + 1) dx = [x^3 - 2x^2 + x]_0^2\)
\(= F(2) - F(0)\)
\(= [2^3 - 2(2)^2 + 2] - [0^3 - 2(0)^2 + 0]\)
\(= [8 - 8 + 2] - [0]\)
\(= 2 - 0 = 2\)
Result: \(\int_0^2 (3x^2 - 4x + 1) dx = 2\)
Compute: \(\int_{0}^{30^\circ} (\sin x + \cos 2x) dx\)
Solution:
First, find the antiderivative:
\(\int (\sin x + \cos 2x) dx = -\cos x + \frac{1}{2} \sin 2x + C\)
Now apply FTC II:
\(\int_{0}^{30^\circ} (\sin x + \cos 2x) dx = [-\cos x + \frac{1}{2} \sin 2x]_{0}^{30^\circ}\)
\(= [-\cos 30^\circ + \frac{1}{2} \sin 60^\circ] - [-\cos 0^\circ + \frac{1}{2} \sin 0^\circ]\)
\(= [-\frac{\sqrt{3}}{2} + \frac{1}{2} \cdot \frac{\sqrt{3}}{2}] - [-1 + 0]\)
\(= [-\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{4}] + 1\)
\(= [-\frac{\sqrt{3}}{4}] + 1\)
\(= 1 - \frac{\sqrt{3}}{4}\)
Result: \(\int_{0}^{30^\circ} (\sin x + \cos 2x) dx = 1 - \frac{\sqrt{3}}{4}\)
Compute: \(\int_{0}^{1} 2x\sqrt{x^2+1} dx\) using algebraic substitution
Solution:
Let \( u = x^2 + 1 \), then \( du = 2x dx \)
When \( x = 0 \), \( u = 0^2 + 1 = 1 \)
When \( x = 1 \), \( u = 1^2 + 1 = 2 \)
Substitute:
\(\int_{0}^{1} 2x\sqrt{x^2+1} dx = \int_{1}^{2} \sqrt{u} du\)
\(= \int_{1}^{2} u^{1/2} du\)
\(= \left[\frac{2}{3} u^{3/2}\right]_{1}^{2}\)
\(= \frac{2}{3} (2^{3/2}) - \frac{2}{3} (1^{3/2})\)
\(= \frac{2}{3} (2\sqrt{2} - 1)\)
Result: \(\int_{0}^{1} 2x\sqrt{x^2+1} dx = \frac{2}{3}(2\sqrt{2} - 1)\)
Properties of Definite Integrals
Constant multiple property
Sum property
Difference property
Reversal of limits property
Same limits property
Additivity property (\(a < c < b\))
Horizontal shift property
If \(\int_1^b (2x - 3)dx = 12\) with \(b > 0\), find the value of \(b\)
Solution:
\(\int_1^b (2x - 3)dx = 12\)
\([x^2 - 3x]_1^b = 12\)
\((b^2 - 3b) - (1^2 - 3 \cdot 1) = 12\)
\((b^2 - 3b) - (1 - 3) = 12\)
\(b^2 - 3b + 2 = 12\)
\(b^2 - 3b - 10 = 0\)
Solve the quadratic equation:
\((b + 2)(b - 5) = 0\)
\(b = -2\) or \(b = 5\)
Since \(b > 0\), the solution is \(b = 5\)
Result: \(b = 5\)
Given \(\int_0^3 f(x)dx = -4\) and \(\int_0^9 f(x)dx = 16\), find \(\int_3^9 f(x)dx\)
Solution: Using Property 6 (Additivity):
\(\int_0^9 f(x)dx = \int_0^3 f(x)dx + \int_3^9 f(x)dx\)
\(16 = -4 + \int_3^9 f(x)dx\)
\(\int_3^9 f(x)dx = 16 + 4 = 20\)
Result: \(\int_3^9 f(x)dx = 20\)
If \(\int_0^7 (x + 5)dx = a\), express \(\int_{1000}^{1007} (2x - 2005)dx\) in terms of \(a\)
Solution: Using Property 7 (Horizontal shift):
Let's shift by subtracting 1000 from the limits:
\(\int_{1000}^{1007} (2x - 2005)dx = \int_{0}^{7} (2(x + 1000) - 2005)dx\)
\(= \int_{0}^{7} (2x + 2000 - 2005)dx\)
\(= \int_{0}^{7} (2x - 5)dx\)
Now manipulate to get the given form:
\(2x - 5 = 2(x + 5) - 15\)
So: \(\int_{0}^{7} (2x - 5)dx = \int_{0}^{7} [2(x + 5) - 15]dx\)
\(= 2\int_{0}^{7} (x + 5)dx - \int_{0}^{7} 15dx\)
\(= 2a - [15x]_{0}^{7}\)
\(= 2a - (15 \cdot 7 - 15 \cdot 0)\)
\(= 2a - 105\)
Result: \(\int_{1000}^{1007} (2x - 2005)dx = 2a - 105\)
The velocity function of an object is: \( V(t) = \begin{cases} 3t & \text{if } 0 \leq t \leq 1 \\ 3 & \text{if } 1 \leq t \leq 6 \end{cases} \)
Assuming the object is at point (0,0) when \(t = 0\), find its position at \(t = 5\)
Solution: Position is the integral of velocity
\(s(5) = \int_{0}^{5} V(t)dt\)
Since \(V(t)\) is piecewise, split the integral:
\(= \int_{0}^{1} 3t dt + \int_{1}^{5} 3 dt\)
Compute each part:
\(\int_{0}^{1} 3t dt = \left[\frac{3}{2}t^2\right]_{0}^{1} = \frac{3}{2}(1)^2 - \frac{3}{2}(0)^2 = \frac{3}{2}\)
\(\int_{1}^{5} 3 dt = [3t]_{1}^{5} = 3(5) - 3(1) = 15 - 3 = 12\)
Add them together:
\(s(5) = \frac{3}{2} + 12 = \frac{3}{2} + \frac{24}{2} = \frac{27}{2} = 13.5\)
Result: The object's position at \(t = 5\) is \(13.5\) units from the starting point
Key Takeaways
Indefinite Integral: \(\int f(x)dx = F(x) + C\) (family of functions)
Definite Integral: \(\int_a^b f(x)dx = F(b) - F(a)\) (specific number)
1. Linearity: \(\int_a^b [kf(x) \pm mg(x)]dx = k\int_a^b f(x)dx \pm m\int_a^b g(x)dx\)
2. Additivity: \(\int_a^b f(x)dx = \int_a^c f(x)dx + \int_c^b f(x)dx\) (for \(a < c < b\))
3. Reversal: \(\int_a^b f(x)dx = -\int_b^a f(x)dx\)
4. Zero length: \(\int_a^a f(x)dx = 0\)
For problems involving properties of definite integrals (like Examples 4-6 above), you must use the properties. These types of problems may seem challenging, especially forms like the examples above. With frequent practice and careful attention, we believe you can master this material well and correctly.