Suppose that ∫ 0 4 f ( x ) d x = 5 and ∫ 0 2 f ( x ) d x = − 3 , and ∫ 0 4 g ( x ) d x = − 1 and ∫ 0 2 g ( x ) d x = 2 . In the following exercises, compute the integrals. 92. ∫ 0 2 ( 3 f ( x ) − 4 g ( x ) ) d x
Suppose that ∫ 0 4 f ( x ) d x = 5 and ∫ 0 2 f ( x ) d x = − 3 , and ∫ 0 4 g ( x ) d x = − 1 and ∫ 0 2 g ( x ) d x = 2 . In the following exercises, compute the integrals. 92. ∫ 0 2 ( 3 f ( x ) − 4 g ( x ) ) d x
Suppose that
∫
0
4
f
(
x
)
d
x
=
5
and
∫
0
2
f
(
x
)
d
x
=
−
3
, and
∫
0
4
g
(
x
)
d
x
=
−
1
and
∫
0
2
g
(
x
)
d
x
=
2
. In the following exercises, compute the integrals.
92.
∫
0
2
(
3
f
(
x
)
−
4
g
(
x
)
)
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.
By considering appropriate series expansions,
e². e²²/2. e²³/3.
....
=
= 1 + x + x² + ·
...
when |x| < 1.
By expanding each individual exponential term on the left-hand side
the coefficient of x- 19 has the form
and multiplying out,
1/19!1/19+r/s,
where 19 does not divide s. Deduce that
18! 1 (mod 19).
Proof: LN⎯⎯⎯⎯⎯LN¯ divides quadrilateral KLMN into two triangles. The sum of the angle measures in each triangle is ˚, so the sum of the angle measures for both triangles is ˚. So, m∠K+m∠L+m∠M+m∠N=m∠K+m∠L+m∠M+m∠N=˚. Because ∠K≅∠M∠K≅∠M and ∠N≅∠L, m∠K=m∠M∠N≅∠L, m∠K=m∠M and m∠N=m∠Lm∠N=m∠L by the definition of congruence. By the Substitution Property of Equality, m∠K+m∠L+m∠K+m∠L=m∠K+m∠L+m∠K+m∠L=°,°, so (m∠K)+ m∠K+ (m∠L)= m∠L= ˚. Dividing each side by gives m∠K+m∠L=m∠K+m∠L= °.°. The consecutive angles are supplementary, so KN⎯⎯⎯⎯⎯⎯∥LM⎯⎯⎯⎯⎯⎯KN¯∥LM¯ by the Converse of the Consecutive Interior Angles Theorem. Likewise, (m∠K)+m∠K+ (m∠N)=m∠N= ˚, or m∠K+m∠N=m∠K+m∠N= ˚. So these consecutive angles are supplementary and KL⎯⎯⎯⎯⎯∥NM⎯⎯⎯⎯⎯⎯KL¯∥NM¯ by the Converse of the Consecutive Interior Angles Theorem. Opposite sides are parallel, so quadrilateral KLMN is a parallelogram.
By considering appropriate series expansions,
ex · ex²/2 . ¸²³/³ . . ..
=
= 1 + x + x² +……
when |x| < 1.
By expanding each individual exponential term on the left-hand side
and multiplying out, show that the coefficient of x 19 has the form
1/19!+1/19+r/s,
where 19 does not divide s.
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