Problem 18 - Applied Calculus

Question

Using the Fundamental Theorem, evaluate the definite integrals in Problems \(1-20\) exactly. $$\int_{0}^{\pi / 4}(\sin t+\cos t) d t$$

Step-by-Step Solution

Answer

The integral evaluates to 1.

Step 1: Identify the Integral Components

The integral given is \( \int_{0}^{\pi / 4} (\sin t + \cos t) \, dt \). Here, you need to integrate both \( \sin t \) and \( \cos t \) over the interval \([0, \pi/4]\).

Step 2: Find the Antiderivative

The antiderivative of \( \sin t \) is \(-\cos t\), and the antiderivative of \( \cos t \) is \( \sin t\). Hence, the antiderivative of \( \sin t + \cos t \) is \(-\cos t + \sin t\).

Step 3: Apply the Fundamental Theorem of Calculus

According to the Fundamental Theorem of Calculus, evaluate the antiderivative at the upper limit, \( \pi/4 \), and the lower limit, \( 0 \), and find the difference. Compute \( \left[ -\cos t + \sin t \right]_{0}^{\pi/4} \).

Step 4: Compute the Antiderivative at the Upper Limit

Substitute \( t = \pi/4 \) into the antiderivative to get \(-\cos(\pi/4) + \sin(\pi/4) = -\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}\).

Step 5: Compute the Antiderivative at the Lower Limit

Substitute \( t = 0 \) into the antiderivative to get \(-\cos(0) + \sin(0) = -1 + 0 = -1\).

Step 6: Evaluate the Definite Integral

Subtract the result at the lower limit from that at the upper limit: \((-\cos(\pi/4) + \sin(\pi/4)) - (-\cos(0) + \sin(0)) = (0) - (-1) = 1\).

Key Concepts

Definite Integrals

Definite integrals are fundamental tools in calculus. They are used to calculate the accumulation of quantities, such as area under a curve, over a specific interval. The definite integral of a function, denoted as \(\int_{a}^{b} f(t)dt\), is computed over the interval \([a, b]\). In this interval, the function \(f(t)\) is integrated from \(a\) to \(b\).

The process involves finding an antiderivative (or indefinite integral) of the function, which is a new function that represents the accumulated area under \(f(t)\) from a starting point up to any point \(t\).

• Steps: First, find an antiderivative of the given function.
• Next, evaluate this antiderivative at the upper and lower limits of the interval.
• Finally, subtract the value obtained at the lower limit from the value at the upper limit.

These steps are supported by the Fundamental Theorem of Calculus, which bridges the concept of derivatives and integrals, allowing us to compute exact values for definite integrals.

Antiderivative

An antiderivative, also known as an indefinite integral, is a function whose derivative is the given function. Finding an antiderivative is one of the essential tasks in calculus, particularly when calculating definite integrals.

To solve integrals like \(\int (\sin t + \cos t) dt\), start by determining the antiderivative of each part of the expression:

• For \(\sin t\), the antiderivative is \(-\cos t\).
• For \(\cos t\), the antiderivative is \(\sin t\).

Thus, the antiderivative of \(\sin t + \cos t\) becomes \(-\cos t + \sin t\).

Once computed, this antiderivative is used to evaluate the definite integral over a given interval \([a, b]\). By calculating the difference \([F(b) - F(a)]\), where \(F\) represents the antiderivative, we obtain the accumulated value or the total change, which is the essence of a definite integral.

Trigonometric Functions

Trigonometric functions like \(\sin t\) and \(\cos t\) appear frequently in calculus problems. Understanding these functions and their properties is crucial when dealing with integrals.

These functions are periodic, meaning they repeat their values in regular intervals. This property is often utilized in integration, especially in problems involving definite integrals.

• Sine Function: \(\sin t\) starts at 0, increases to 1, decreases to -1, and repeats.
• Cosine Function: \(\cos t\) starts at 1, decreases to 0, decreases further to -1, and repeats.

Additionally, certain angles such as \(\pi/4\), \(\pi/2\), \(\pi\), and 0 have known sine and cosine values which simplify integration computations:

• \(\sin(\pi/4) = \cos(\pi/4) = \frac{\sqrt{2}}{2}\)
• \(\sin(0) = 0\), \(\cos(0) = 1\)

Recognizing these values aids in quickly evaluating integrals like the one in the original exercise, where these functions are combined and integrated over specific intervals.