Integration by parts: definite integrals | AP Calculus BC | Khan Academy

Khan Academy
1 Aug 201705:42
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TLDRThe video script discusses the process of evaluating a specific definite integral using integration by parts. The integral in question is from zero to pi of x times the cosine of x. The instructor explains that direct anti-differentiation is not straightforward and suggests using integration by parts for simplification. The choice of functions for integration by parts is discussed, with x chosen as f(x) and cosine of x as g'(x), leading to the anti-derivative of cosine x, sine x. The final calculation involves evaluating the result at the bounds and subtracting, resulting in a final answer of negative two.

Takeaways
  • ๐Ÿ“š The video focuses on evaluating a specific definite integral: โˆซ from 0 to ฯ€ of x*cos(x) dx.
  • ๐Ÿค” This integral cannot be easily evaluated by simply taking the antiderivative due to the product of x and cos(x).
  • ๐ŸŽ“ A more sophisticated technique is required, hinting at the use of integration by parts for this problem.
  • ๐Ÿ“ˆ Integration by parts is a method that utilizes the product rule from differential calculus to simplify integrals involving a product of two functions.
  • ๐Ÿ”„ The formula for integration by parts is โˆซ u dv = uv - โˆซ v du, where u and dv are chosen to simplify the resulting expressions.
  • ๐Ÿฆ  The choice of u and dv is crucial: u should be the function that simplifies when differentiated, and dv should simplify when integrated.
  • ๐ŸŒŸ In this case, x is chosen as u (since its derivative is simpler, just 1) and cos(x) as dv (since its integral, sine(x), does not complicate further).
  • ๐Ÿ“Š Applying integration by parts, the integral becomes x*sin(x) - โˆซ sin(x) dx, which simplifies further.
  • ๐Ÿงฎ The indefinite integral of sin(x) is -cos(x), so the expression now is x*sin(x) + cos(x).
  • ๐Ÿ To find the definite integral, evaluate the expression at the bounds (ฯ€ and 0) and subtract the lower from the upper.
  • ๐ŸŽฏ After evaluation, the result of the integral is -2, showcasing the effectiveness of integration by parts in solving the problem.
Q & A

  • What is the main topic of the video?

    -The main topic of the video is evaluating the definite integral from zero to pi of x cosine of x dx using integration by parts.

  • Why is it not straightforward to evaluate the given integral using the basic anti-derivative method?

    -The given integral does not allow for a straightforward evaluation using the basic anti-derivative method because the product of functions, x and cosine of x, complicates the process. The derivative of x is simple (one), but the anti-derivative of cosine of x does not simplify the expression further.

  • What is integration by parts and how does it help in this scenario?

    -Integration by parts is a technique used in calculus to evaluate integrals when a direct approach is not feasible. It is based on the product rule from differential calculus and helps simplify the process by separating the integral into more manageable parts. In this scenario, it helps because one of the functions (x) simplifies when differentiated, and the other (cosine of x) does not complicate when integrated, making the evaluation process easier.

  • How do you choose the functions f(x) and g'(x) for integration by parts?

    -In integration by parts, you choose f(x) to be the function that simplifies when you take its derivative and g'(x) to be the function that does not complicate when you take its anti-derivative. In this case, x is chosen as f(x) because its derivative is one, and cosine of x is chosen as g'(x) because its anti-derivative, sine of x, does not become more complicated.

  • What is the indefinite integral of the function g'(x) = cosine of x?

    -The indefinite integral of the function g'(x) = cosine of x is g(x) = sine of x. This is because the derivative of sine of x is cosine of x, which matches the given g'(x) function.

  • How does the process of integration by parts simplify the original integral?

    -Integration by parts simplifies the original integral by breaking it down into a more manageable form. In this case, the integral becomes x times sine of x minus the integral of sine of x dx, which further simplifies to x sine of x minus negative cosine of x, as the anti-derivative of sine of x is negative cosine of x.

  • What are the steps to evaluate the integral at the endpoints using integration by parts?

    -First, you evaluate the product of the chosen f(x) and g(x) at the endpoints (pi and zero). Then, you evaluate the integral of the second part (the one involving the derivative of f(x)) at the endpoints. Finally, you combine these evaluations, taking the difference between the endpoint evaluations to find the definite integral.

  • What is the value of sine of pi and cosine of pi?

    -The value of sine of pi is zero, and the value of cosine of pi is negative one.

  • What is the final result of the definite integral after applying integration by parts?

    -The final result of the definite integral after applying integration by parts is negative two.

  • How does the process of integration by parts help in understanding complex integrals?

    -Integration by parts provides a systematic approach to breaking down complex integrals into simpler components, making them easier to evaluate. It helps in understanding the relationship between the functions being integrated and their derivatives and anti-derivatives, which is crucial in advanced calculus.

  • What is the significance of remembering that arbitrary constants cancel out in definite integrals?

    -The significance of remembering that arbitrary constants cancel out in definite integrals is that it simplifies the process of integration. Since the constants are not part of the function's behavior over the interval of integration, they do not affect the result of the integral and can be ignored, making the calculations cleaner and more straightforward.

Outlines
00:00
๐Ÿ“š Introduction to Integration by Parts

The first paragraph introduces the concept of evaluating a definite integral using a more sophisticated technique, specifically integration by parts. The integral in question involves the product of x and the cosine of x. The instructor explains that when faced with a product of functions, one should look for a function that simplifies when differentiated (in this case, x) and another that does not complicate when integrated (cosine of x). The explanation sets the stage for applying the integration by parts formula, which is derived from the product rule in differential calculus. The goal is to find f(x) and g'(x) such that the integral of their product can be simplified. The choice of f(x) as x and g'(x) as cosine of x is justified, leading to the anti-derivative g(x) being sine of x.

05:01
๐Ÿ”ข Completion of Integration by Parts

The second paragraph details the completion of the integration by parts process. The instructor applies the integration by parts formula to the given integral, resulting in the expression x times the sine of x minus the integral of the product of the derivative of x (which is 1) and the sine of x. This simplifies to x times the sine of x minus the integral of the sine of x. The next step is to evaluate this expression at the endpoints, pi and zero. The evaluation at pi results in pi times the sine of pi (which is zero) plus the cosine of pi (which is -1). The evaluation at zero results in zero times the sine of zero (which is zero) plus the cosine of zero (which is 1). Subtracting the value evaluated at zero from that at pi yields the final result of negative two, thus completing the evaluation of the definite integral using integration by parts.

Mindmap
Keywords
๐Ÿ’กDefinite Integral
A definite integral represents the area under a curve of a function over a specified interval. In the video, the definite integral is used to calculate the area under the curve of the function x*cos(x) from 0 to ฯ€, which is the main problem being solved.
๐Ÿ’กAnti-Derivative
The anti-derivative is a function that, when differentiated, yields the original function. In the context of the video, finding the anti-derivative of a function is essential for evaluating the definite integral, as it allows us to apply integration techniques.
๐Ÿ’กIntegration by Parts
Integration by parts is a technique used to evaluate integrals when the integrand is a product of two functions. In the video, integration by parts is chosen as the method to evaluate the integral because the product of x and cos(x) suggests that this approach will simplify the calculation.
๐Ÿ’กProduct Rule
The product rule is a fundamental calculus concept that describes how to differentiate the product of two functions. Although the product rule is not directly used to evaluate the integral in the video, the concept is related to integration by parts, which is based on the product rule from differential calculus.
๐Ÿ’กSine and Cosine Functions
Sine and cosine are trigonometric functions that are often used in calculus, especially in integration. In the video, the integral involves the product of x and cos(x), and the anti-derivative of cos(x) is sin(x), which plays a key role in applying integration by parts.
๐Ÿ’กDerivative
The derivative of a function gives the rate of change or the slope of the function at any point. In the video, the derivatives of the functions involved in the integral are considered to determine the best application of integration by parts.
๐Ÿ’กArbitrary Constants
In calculus, an arbitrary constant is often introduced when finding an anti-derivative. However, in the context of definite integrals, as mentioned in the video, these constants cancel out because we are interested in the difference between the values of the anti-derivative at the endpoints of the integration interval.
๐Ÿ’กEvaluation at Endpoints
When dealing with definite integrals, it is necessary to evaluate the anti-derivative at the endpoints of the integration interval. In the video, the evaluation at ฯ€ and at 0 is performed to find the area under the curve of x*cos(x) between these two points.
๐Ÿ’กArea under a Curve
The area under a curve represents the accumulated value over the interval for a given function. In the video, the goal is to find this area for the function x*cos(x) between the limits 0 and ฯ€, which is a fundamental concept in integral calculus.
๐Ÿ’กSimplification
Simplification in mathematics refers to making a complex expression or equation more straightforward. In the video, the process of simplifying the integrand by choosing appropriate functions for f(x) and g(x) in the integration by parts formula is crucial for effectively evaluating the integral.
๐Ÿ’กNegative Results
In the context of the video, the negative result obtained from the evaluation of the definite integral (negative two) is an example of how integrals can yield positive or negative values depending on the function and the interval. This illustrates the importance of correctly evaluating the anti-derivative at the endpoints.
Highlights

The video aims to evaluate the definite integral from zero to pi of x times the cosine of x dx.

It's not straightforward to evaluate the integral using basic anti-derivative methods.

Integration by parts is suggested as a more sophisticated technique for this problem.

Integration by parts is applicable when there's a product of functions, one of which simplifies upon differentiation and the other upon integration.

The formula for integration by parts is provided, which is an application of the product rule from differential calculus.

For the given integral, x is chosen as f(x) because its derivative simplifies to 1.

The function cosine of x is chosen as g'(x) because its anti-derivative does not complicate further.

The anti-derivative of cosine of x is sine of x, and the derivative of x is 1.

The integration by parts formula is applied to the given integral, resulting in x times sine of x minus the integral of sine of x.

The integral is evaluated at pi and zero, and the results are subtracted to find the definite integral.

The anti-derivative of sine of x is cosine of x, which simplifies the evaluation process.

The evaluation at pi results in pi times sine of pi plus cosine of pi.

The evaluation at zero results in zero times sine of zero plus cosine of zero.

Sine of pi equals zero and cosine of zero equals one, leading to a simplified expression.

Cosine of pi equals negative one, and combining the evaluations results in a final answer of negative two.

The method of integration by parts successfully evaluates the given definite integral.

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