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$$\text { Show that } \frac{1}{\sqrt{2 \pi}} \int_{-1.5}^{2} e^{-\frac{e^{2}}{2}} d t=\frac{2}{\sqrt{2 \pi}} \int_{0}^{1.5} e^{-\frac{e^{2}}{2}} d t+\frac{1}{\sqrt{2 \pi}} \int_{1.5}^{2} e^{-\frac{e^{2}}{2}} d t$$

$$\pm 0.674537$$

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 7

Substitution and Properties of Definite Integrals

Integrals

Campbell University

Harvey Mudd College

Baylor University

University of Nottingham

Lectures

05:53

In mathematics, an indefinite integral is an integral whose integrand is not known in terms of elementary functions. An indefinite integral is usually encountered when integrating functions that are not elementary functions themselves.

40:35

In mathematics, integration is one of the two main operations of calculus, with its inverse operation, differentiation, being the other. Given a function of a real variable (often called "the integrand"), an antiderivative is a function whose derivative is the given function. The area under a real-valued function of a real variable is the integral of the function, provided it is defined on a closed interval around a given point. It is a basic result of calculus that an antiderivative always exists, and is equal to the original function evaluated at the upper limit of integration.

00:48

Show that $\int_{-a}^{a} e…

01:22

Show that

So the question says soul integral minus 8 to 8 is super minus U. Squared du because route by er F. F. So let's hold this question. So let's hold this question. So we have we have since therefore facts first too. It is to the power minus X. So scared this isn't and is in you in anyone. This is an even function. So integral minus A. To a. It is two part of mine issues here. Do you? First two tries oh, integer zero to a. It has to call minus you scared you? Therefore minus two integral. It is super minus you scared do you? Because brought by bye. Yeah. Okay.

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