Problem 183
Question
In the following exercises, evaluate each definite integral using the Fundamental Theorem of Calculus, Part 2 . $$ \int_{0}^{\pi / 2} \sin \theta d \theta $$
Step-by-Step Solution
Verified Answer
The value of the integral is 1.
1Step 1: Identify the Integral Limits
The given definite integral is \( \int_{0}^{\pi/2} \sin \theta \, d\theta \). The lower limit is 0 and the upper limit is \( \pi/2 \).
2Step 2: Find the Antiderivative
Use the Fundamental Theorem of Calculus, which states that if \( F \) is the antiderivative of \( f \), then \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \). The antiderivative of \( \sin \theta \) is \( -\cos \theta \).
3Step 3: Apply the Limits to the Antiderivative
Evaluate the antiderivative \(-\cos \theta\) at the upper limit \( \pi/2 \) and the lower limit 0. This gives \(-\cos(\pi/2) - (-\cos(0))\).
4Step 4: Calculate the Values
Substitute the values into the expression: \(-\cos(\pi/2)\) is 0 (since \(\cos(\pi/2) = 0\)) and \(-\cos(0)\) is -1 (since \(\cos(0) = 1\)).
5Step 5: Simplify the Result
The calculation becomes \(0 - (-1) = 1\). Thus, \( \int_{0}^{\pi/2} \sin \theta \, d\theta = 1 \).
Key Concepts
Definite IntegralAntiderivativeTrigonometric Integrals
Definite Integral
A definite integral can be thought of as the area under a curve within a specified interval. In this exercise, we evaluate the definite integral \(\int_{0}^{\pi/2} \sin \theta \, d\theta\). This integral has a specific lower limit \(0\) and an upper limit \(\pi/2\).
To solve for this, we use the second part of the Fundamental Theorem of Calculus. The theorem tells us that if we know an antiderivative of a function, we can find the area under the curve by evaluating this antiderivative at the upper and lower limits of the integral and subtracting the two values.
Here’s why the definite integral is so practical:
To solve for this, we use the second part of the Fundamental Theorem of Calculus. The theorem tells us that if we know an antiderivative of a function, we can find the area under the curve by evaluating this antiderivative at the upper and lower limits of the integral and subtracting the two values.
Here’s why the definite integral is so practical:
- It efficiently computes the accumulated total (like area, distance, etc.) over an interval.
- This concept is widely used in physics and engineering for calculating things like work, energy, or electric charge over a distance.
Antiderivative
An antiderivative of a function \(f\) is another function \(F\) whose derivative is \(f\). This means, if you differentiate \(F\), you will get back the original function \(f\).
In our problem, we're working with \(\sin \theta\). The antiderivative of \(\sin \theta\) is \(-\cos \theta\). Why is this the case? Because the derivative of \(-\cos \theta\) is \(\sin \theta\).
Here’s a quick reminder about antiderivatives:
In our problem, we're working with \(\sin \theta\). The antiderivative of \(\sin \theta\) is \(-\cos \theta\). Why is this the case? Because the derivative of \(-\cos \theta\) is \(\sin \theta\).
Here’s a quick reminder about antiderivatives:
- The process of finding an antiderivative is the reverse of differentiation.
- Antiderivatives are critical for solving definite integrals because of the Fundamental Theorem of Calculus.
Trigonometric Integrals
Trigonometric integrals involve functions like \(\sin\), \(\cos\), \(\tan\), etc. They are a key part of calculus because trigonometric functions appear frequently in physics and engineering problems.
In our example, we evaluated the integral \( \int \sin \theta \, d\theta \) using its antiderivative \(-\cos \theta\). Trigonometric integrals can sometimes seem tricky because of the oscillating nature of these functions, but they follow the same rules as other integrals.
Here are some tips for working with trigonometric integrals:
In our example, we evaluated the integral \( \int \sin \theta \, d\theta \) using its antiderivative \(-\cos \theta\). Trigonometric integrals can sometimes seem tricky because of the oscillating nature of these functions, but they follow the same rules as other integrals.
Here are some tips for working with trigonometric integrals:
- Focus on learning the basic antiderivatives of \(\sin x\), \(\cos x\), \(\tan x\), etc.
- Remember that integrals involving products or powers of trigonometric functions might need special techniques (like substitution).
Other exercises in this chapter
Problem 181
In the following exercises, evaluate each definite integral using the Fundamental Theorem of Calculus, Part 2 . $$ \int_{1}^{16} \frac{d t}{t^{1 / 4}} $$
View solution Problem 182
In the following exercises, evaluate each definite integral using the Fundamental Theorem of Calculus, Part 2 . $$ \int_{0}^{2 \pi} \cos \theta d \theta $$
View solution Problem 184
In the following exercises, evaluate each definite integral using the Fundamental Theorem of Calculus, Part 2 . $$ \int_{0}^{\pi / 4} \sec ^{2} \theta d \theta
View solution Problem 185
In the following exercises, evaluate each definite integral using the Fundamental Theorem of Calculus, Part 2 . \(\int_{0}^{\pi / 4} \sec \theta \tan \theta\)
View solution