Using Properties of Definite Integrals In Exercises 37-44, evaluate the definite integral using the values below. ∫ 2 6 x 3 d x = 320 , ∫ 2 6 x d x = 16 , ∫ 2 6 d x = 4 ∫ 2 6 ( 2 x 3 − 10 + 7 ) d x
Using Properties of Definite Integrals In Exercises 37-44, evaluate the definite integral using the values below. ∫ 2 6 x 3 d x = 320 , ∫ 2 6 x d x = 16 , ∫ 2 6 d x = 4 ∫ 2 6 ( 2 x 3 − 10 + 7 ) d x
Solution Summary: The author explains how to calculate a definite integral displaystyle, using the provided values.
Using Properties of Definite Integrals In Exercises 37-44, evaluate the definite integral using the values below.
∫
2
6
x
3
d
x
=
320
,
∫
2
6
x
d
x
=
16
,
∫
2
6
d
x
=
4
∫
2
6
(
2
x
3
−
10
+
7
)
d
x
With differentiation, one of the major concepts of calculus. Integration involves the calculation of an integral, which is useful to find many quantities such as areas, volumes, and displacement.
Complex analysis: Solve this integral using contour integration.
PLS SOLVE INTEGRAL CALCULUS
Evaluating a Definite Integral In Exercises
43-52, evaluate the definite integral. Use a
graphing utility to verify your result.
1/4
45.
x cos 2x dx
46.
x sin 2x dx
*n/8
51.
х arcsec x dx
52.
x sec² 2x dx
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Definite Integral Calculus Examples, Integration - Basic Introduction, Practice Problems; Author: The Organic Chemistry Tutor;https://www.youtube.com/watch?v=rCWOdfQ3cwQ;License: Standard YouTube License, CC-BY