If f is continuous on the closed interval [a,b] and F′ = f, then, according to the Fundamental Theorem of Calculus
The following theorems about definite integrals are important.
Example:
∫2−1(3x2−2x) dx=x3−x2|2−1=(8−4)−(−1−1)=6
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This theorem provides the tool for evaluating an infinite sum by means of a definite integral.
Δxk is called a Riemann Sum.
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Example:
Solution:
If f(x) is nonnegative on [a,b], we see (Figure 6–1) that f(xk) Δx can be regarded as the area of a typical approximating rectangle.
Assuming equal width rectangles, as the number of rectangles increases, or, equivalently, as the width Δx of the rectangles approaches zero, the rectangles become an increasingly better fit to the curve.
The sum of their areas gets closer and closer to the exact area under the curve.
Finally, the area bounded by the x-axis, the curve, and the vertical lines x = a and x = b is given exactly by
What if f(x) is negative? Then any area above the graph and below the x-axis is counted as negative (Figure 6–2).
Geometrically, area is always positive, so the shaded area above the curve and below the x-axis equals
where the integral yields a negative number. Note that every product f(xk) Δx in the shaded region is negative, since f(xk) is negative for all x between a and b.
We may approximate ∫b a f(x) dx by any of the following sums, where Δx represents the subinterval widths:
(1)Left sum: f(x0) Δx1 + f(x1) Δx2 + ⋯ + f(xn − 1) Δxn, using the value of f at the left endpoint of each subinterval.
(2)Right sum: f(x1) Δx1 + f(x2) Δx2 + ⋯ + f(xn) Δxn, using the value of f at the right end of each subinterval.
(3)Midpoint sum:
using the value of f at the midpoint of each subinterval.
Example :
Approximate ∫20x3 dx by using four subintervals of equal width and calculating:
(a)the left sum
(b)the right sum
(c)the midpoint sum
(d)the integral
SOLUTIONS:
We now find the areas of the strips in Figure 6–6 by using trapezoids. We denote the bases of the trapezoids by y0*, y1, y2, . . ., yn* and the heights by Δx = h1, h2, . . ., hn.
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SOLUTION:
h=12.ℎ=12. Then,
From Figure 6–7 we infer that the area of a trapezoid is less than the true area if the graph of f is concave down, but is more than the true area if the graph of f is concave up
Figure 6–8 is helpful in showing how the area of a midpoint rectangle compares with that of a trapezoid and with the true area.
Our graph here is concave down. If M is the midpoint of AB, then the midpoint rectangle is AM1M2B. We’ve drawn T1T2 tangent to the curve at T (where the midpoint ordinate intersects the curve).
Generalizing to n subintervals, we conclude:
Example:
Write an inequality including L(n), R(n), M(n), T(n), and ∫baf(t) dtfor the graph of f shown in Figure 6–9.
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It is quite common to define ln x, the natural logarithm of x, as a definite integral, as follows:
This integral can be interpreted as the area bounded above by the curve y=1t(t>0), below by the t-axis, at the left by t = 1, and at the right by t = x (x > 1).
Note that if x = 1 the above definition yields ln 1 = 0, and if 0 < x < 1 we can rewrite as follows:
showing that ln x < 0 if 0 < x < 1.
With this definition of ln x we can approximate ln x using rectangles or trapezoids.
Example :
Show that 12<ln 2<1.12<ln 2<1.
SOLUTION:
If the function y = f(x) is integrable on the interval a ⩽ x ⩽ b, then we define the average value of f from a to b to be
Note that (1) is equivalent to
If f(x) ⩾ 0 for all x on [a,b], we can interpret (2) in terms of areas as follows:
Example :
Find the average value of f(x) = ln x on the interval [1,4].
SOLUTION:
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