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Evaluate the integral.

$ \displaystyle \int z^3 e^z dz $

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$I_{1}=z^{3} e^{z}-3\left(z^{2} e^{z}-2 z e^{z}+2 e^{z}-2 C_{1}\right)=z^{3} e^{z}-3 z^{2} e^{z}+6 z e^{z}-6 e^{z}+C,$ where $C=6 C_{1}$

Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 1

Integration by Parts

Integration Techniques

Missouri State University

Harvey Mudd College

University of Michigan - Ann Arbor

Idaho State University

Lectures

01:53

In mathematics, integration is one of the two main operations in calculus, with its inverse, differentiation, being the other. Given a function of a real variable, an antiderivative, integral, or integrand is the function's derivative, with respect to the variable of interest. The integrals of a function are the components of its antiderivative. The definite integral of a function from a to b is the area of the region in the xy-plane that lies between the graph of the function and the x-axis, above the x-axis, or below the x-axis. The indefinite integral of a function is an antiderivative of the function, and can be used to find the original function when given the derivative. The definite integral of a function is a single-valued function on a given interval. It can be computed by evaluating the definite integral of a function at every x in the domain of the function, then adding the results together.

27:53

In mathematics, a technique is a method or formula for solving a problem. Techniques are often used in mathematics, physics, economics, and computer science.

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Evaluate the integrals.

01:56

The problem is evaluated, the integral 82 serious power times e to z, d z. For this problem we will use the method of integration by powers. The formula is integral of: u times, prime x is equal to: u times, v minus integral of? U? Prime! For our problem, we can let, u is equal to for our problem. We can let? U is equal to c to 3 power and the prime is equal to e to the derivative of? U is equal to 3 times t square, and the v is equal to 2 d. Now use this formula. We have this integral is equal to u times v. So this is to the 3 power 2 t minus the integral of 3 times z, squared into z, d z, and for this integration fr this integral. We can also use integration by parts that, u is equal to z, square and prime, is equal to into z, and then your prime is equal to 2 times. T v is equal to into z, so this is equal to 23 power to z minus 3 times? U times v, so this is t square into z, minus the integral of pramtimes v. So this is minus 2 times integral of z, times e to z e z. So this is equal to z, 2, serious power to z, minus 3 times e square into z and plus 6 times integral 2 izz. For this integral we use use substitution. We use integration by parts again so that, u is equal to z and prime, is equal to e to z. Then u, prime, is equal to 1 and v is equal to 2, so this integral is equal to u times v. So this is t times a to z, integral of e t z. So this is minus 2, so i answer should be z, square into z, minus 3 times e to 3 times square times to 0 plus 6 times times e to z, minus 6 times e to times 6 minus 6 times e to z and paste constant number C, this is the answer.

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