Definite integral of trig function | AP Calculus AB | Khan Academy
TLDRThe video script presents a step-by-step explanation of evaluating a definite integral of a sine function. The process involves finding the antiderivative of the function, applying integration properties, and evaluating the result at the given bounds. The integral is simplified by relating the sine function to the derivative of the cosine function, and the final calculation results in the value of negative nine, demonstrating a clear understanding of calculus concepts.
Takeaways
- ๐ The problem involves evaluating a definite integral from 11ฯ/2 to 6ฯ of 9sin(x)dx.
- ๐ The first step is to find the antiderivative of 9sin(x), which can be simplified using integration properties.
- ๐ The antiderivative of sine(x) is the negative cosine(x), which is derived from the relationship between derivatives.
- ๐ค The script suggests multiplying the integral by -1 twice to match the derivative of cosine(x), resulting in -9sin(x).
- ๐งฎ The integral is then evaluated by substituting the bounds 6ฯ and 11ฯ/2 into the antiderivative, which is -9cos(x).
- ๐ The value of cosine(6ฯ) is 1, as any multiple of 2ฯ results in 1, which is a fundamental property of the unit circle.
- ๐ To evaluate cosine(11ฯ/2), the script suggests subtracting multiples of 2ฯ until a recognizable value is obtained, which is 3ฯ/2.
- ๐ The cosine of 3ฯ/2 is 0, as it corresponds to the x-coordinate on the unit circle at that angle.
- ๐ข The final expression for the integral simplifies to 1 - 0, which equals 1, before multiplying by -9.
- ๐ The definite integral evaluates to -9, which is the product of -9 and the result of the evaluation of the antiderivative at the bounds.
Q & A
What is the definite integral being evaluated in the script?
-The definite integral being evaluated is from 11 pi/2 to 6 pi of 9 sin(x) dx.
How does the antiderivative of 9 sin(x) relate to the properties of integration?
-The antiderivative of 9 sin(x) can be simplified using integration properties by recognizing it as 9 times the integral of sin(x), which helps in evaluating the definite integral.
What is the antiderivative of sine of x in terms of cosine function?
-The antiderivative of sine of x is the cosine of x, since the derivative of cosine of x is equal to negative sine of x.
Why is the script multiplying by negative one twice?
-Multiplying by negative one twice is done to maintain the original value while allowing the expression to match the derivative of the cosine function, which simplifies the integration process.
How does the value of cosine change at multiples of two pi?
-The value of cosine at multiples of two pi is equal to one, as it represents points on the unit circle where the x-coordinate is one.
What is the value of cosine at 11 pi/2?
-The value of cosine at 11 pi/2 is zero, as it can be expressed as cosine of (11 pi/2 - 5 pi) which is the same as cosine of 3 pi/2, where the x-coordinate on the unit circle is zero.
How does the unit circle definition of cosine help in understanding the definite integral?
-The unit circle definition of cosine provides a geometric interpretation of the function values, which helps in understanding and evaluating the definite integral by visualizing the points on the unit circle corresponding to the angles.
What is the final result of the definite integral as per the script?
-The final result of the definite integral is negative nine, as it is the product of -9 and the difference in cosine values evaluated at the bounds of the integral (cosine of 6 pi which is 1, minus cosine of 11 pi/2 which is 0).
Why is it important to review the unit circle definition of cosine when dealing with problems like this?
-Reviewing the unit circle definition of cosine is important because it provides a foundational understanding of the behavior of the cosine function at different angles, especially at multiples of pi, which is crucial for evaluating integrals and solving trigonometric problems.
What is the significance of the antiderivative being equal to the cosine function in this context?
-The significance of the antiderivative being equal to the cosine function is that it simplifies the process of evaluating the definite integral by allowing us to use the properties of the cosine function and the geometry of the unit circle to find the difference in cosine values at the bounds of the integral.
How does the process of subtracting multiples of two pi help in simplifying the problem?
-Subtracting multiples of two pi helps in simplifying the problem by reducing the angles to values that can be more easily understood and visualized on the unit circle, which in turn makes it easier to evaluate the trigonometric functions and find the solution.
Outlines
๐ Evaluating a Definite Integral
This paragraph discusses the process of evaluating a definite integral of the function nine sine of x from 11 pi over two to six pi. The voiceover explains the initial step of finding the antiderivative of nine sine of x, using integration properties to simplify the expression. The antiderivative is identified as nine times the integral of sine of x, and the derivative relationship between sine and cosine functions is used to match the given function with the derivative of cosine of x. The integral is then evaluated by substituting the bounds of the integration, resulting in the expression cosine of six pi minus cosine of 11 pi over two. The paragraph concludes with the evaluation of the cosine values at these bounds, leading to the final result of the integral being negative nine.
Mindmap
Keywords
๐กDefinite Integral
๐กAntiderivative
๐กIntegration Properties
๐กDerivatives
๐กUnit Circle
๐กTrigonometric Functions
๐กEvaluation at Bounds
๐กNegative Sign
๐กMultiples of Pi
๐กSine Function
๐กCosine Function
Highlights
The process of evaluating a definite integral involving the sine function is discussed.
The integral is from 11 pi/2 to 6 pi of 9 sine of x dx.
The antiderivative of nine sine of x is sought using integration properties.
The relationship between the derivative of cosine x and sine x is utilized.
A method of multiplying by negative one twice to match the derivative of cosine x is introduced.
The antiderivative is expressed as negative nine times the antiderivative of negative sine of x.
The antiderivative is identified as cosine x.
Evaluation of the antiderivative at the bounds 6 pi and 11 pi/2 is performed.
Cosine of a multiple of two pi is shown to equal one.
The value of cosine of 6 pi is determined to be one based on the unit circle definition.
Cosine of 11 pi/2 is simplified by subtracting a multiple of two pi.
The final value of the cosine function after simplification is cosine of 3 pi/2.
The x-coordinate on the unit circle for 3 pi/2 is zero, leading to a cosine value of zero.
The definite integral is evaluated to one minus zero, resulting in a value of one.
The final result of the definite integral is negative nine, multiplying the value of one by negative nine.
Transcripts
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