CHAPTER 20
The Definite Integral and the
Fundamental Theorem of Calculus
20.1
Evaluate 4 dx by the direct (Riemann) definition of the integral.
Let 2 = x 0
be any partition of [2,5], and let A,JC = x. —*,•_,. Then an
Hence, the integral, which is approximated arbitrarily closely by the approximating sums, must be 12.
20.2
Calculate
by the direct definition of the integral.
Divide [0,1] into n equal subintervals, each of length A,* = l//i. In the /th subinterval, choose x* to be the
right endpoint i/n. Then the approximating sum is
As we make the subdivision finer by letting n —» +»,
is the value of the integral.
the approximating sum approaches | • 1 • 2 = f,
which
20.3
Prove the formula
that was used in the solution of Problem 20.2.
Use induction with respect to n. For n = l. the sum consists of one term (I)
2 = 1. The right side
is (l-2-3)/6=l. Now assume that the formula holds for a given positive integer n. We must prove it
for n + I. Adding (n + I)
2
to both sides of the formula
we have
which is the case of the formula for n + 1.
20.4
Prove the formula 1 + 2 + • • • + n =
Let S = 1 + 2 + • • • + (n - 1) + n. Then we also can write 5 = n + (n - 1) + • • • + 2 + 1. If we add
these two equations column by column, we see that 25 is equal to the number n + I added to itself n times.
Thus, 25 = «(« + !), S=n(n + l)/2.
20.5
Show that
by the direct definition of the integral.
Divide the interval [0, b\ into n equal subintervals of length bin, by the points 0 = x 0 < bin <2bln<- • • <
nbln = x = b. In the ith subinterval choose x* to be the right-hand endpoint Ibln. Then an approximating
sum is
As «—»+«>, the approximating sum approaches
b 12, which is, therefore, the value of the integral.
152
4 dx is
5x
2 dx
= 4(jt,, - x 0 ) = 4(5 - 2) = 4 • 3 = 12.
The Definite Integral and the
Fundamental Theorem of Calculus
20.1
Evaluate 4 dx by the direct (Riemann) definition of the integral.
Let 2 = x 0
be any partition of [2,5], and let A,JC = x. —*,•_,. Then an
Hence, the integral, which is approximated arbitrarily closely by the approximating sums, must be 12.
20.2
Calculate
by the direct definition of the integral.
Divide [0,1] into n equal subintervals, each of length A,* = l//i. In the /th subinterval, choose x* to be the
right endpoint i/n. Then the approximating sum is
As we make the subdivision finer by letting n —» +»,
is the value of the integral.
the approximating sum approaches | • 1 • 2 = f,
which
20.3
Prove the formula
that was used in the solution of Problem 20.2.
Use induction with respect to n. For n = l. the sum consists of one term (I)
2 = 1. The right side
is (l-2-3)/6=l. Now assume that the formula holds for a given positive integer n. We must prove it
for n + I. Adding (n + I)
2
to both sides of the formula
we have
which is the case of the formula for n + 1.
20.4
Prove the formula 1 + 2 + • • • + n =
Let S = 1 + 2 + • • • + (n - 1) + n. Then we also can write 5 = n + (n - 1) + • • • + 2 + 1. If we add
these two equations column by column, we see that 25 is equal to the number n + I added to itself n times.
Thus, 25 = «(« + !), S=n(n + l)/2.
20.5
Show that
by the direct definition of the integral.
Divide the interval [0, b\ into n equal subintervals of length bin, by the points 0 = x 0 < bin <2bln<- • • <
nbln = x = b. In the ith subinterval choose x* to be the right-hand endpoint Ibln. Then an approximating
sum is
As «—»+«>, the approximating sum approaches
b 12, which is, therefore, the value of the integral.
152
4 dx is
5x
2 dx
= 4(jt,, - x 0 ) = 4(5 - 2) = 4 • 3 = 12.