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If $ f $ is a positive function and $ f^{\prime \prime} (x) < 0 $ for $ a \le x \le b $, show that$$ T_n < \int_a^b f(x)\ dx < M_n $$

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Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 7

Approximate Integration

Integration Techniques

Oregon State University

Harvey Mudd College

Idaho State University

Boston College

Lectures

01:53

In mathematics, integration is one of the two main operations in calculus, with its inverse, differentiation, being the other. Given a function of a real variable, an antiderivative, integral, or integrand is the function's derivative, with respect to the variable of interest. The integrals of a function are the components of its antiderivative. The definite integral of a function from a to b is the area of the region in the xy-plane that lies between the graph of the function and the x-axis, above the x-axis, or below the x-axis. The indefinite integral of a function is an antiderivative of the function, and can be used to find the original function when given the derivative. The definite integral of a function is a single-valued function on a given interval. It can be computed by evaluating the definite integral of a function at every x in the domain of the function, then adding the results together.

27:53

In mathematics, a technique is a method or formula for solving a problem. Techniques are often used in mathematics, physics, economics, and computer science.

36:58

If $f$ is a positive funct…

02:08

Show that if $f$ is inte…

03:13

If a function f [a,b] = R …

Okay, so this question wants us to show the following here if F double prime of X is negative. So if f is Khan cave down, then the integral is bounded between the trap is a good approximation and the midpoint approximation. So this could be pretty complicated looking to start, So I'm gonna rewrite it in a different form. So if it's bounded by t even below, that means t even it's less enough of X. But it also means that And then it's bigger than X because bounded by that on the bus So we can prove this by showing that the trap is laid. Rule underestimates and at the same time, the midpoint overestimates. Okay, so think it's best to argue this graphically. So let's pick our favorite con cave down curve. It's in this case, just negative, X squared of some variety. I'll just shift it over so we can see what's going on. Better. And let's pick a very large trap is Lloyd So we really see what's going on here. So let's just pretend we have a trap is laid starting here, and I think they're sore. Trap is Lloyd starts here and ends here and then we connect the two segments. So this is true for any trap is laid sizing pick. But as you can see here, this is the area we estimate I'll use a different color. So this is the area? Yes, to May. But we're missing all this black area. So if we consistently miss area under the curve, that means that their trap is Lloyd underestimates. Okay, so now we're halfway done. We've shown that the trap is I'd rule will underestimate the area of the curve. Now let's tackle the midpoint approximation again. Let's do the same curve. But this works with any con cave down. So let's say we have some rectangle and we're taking the mid point value right here and extending it from that X value to that X value. And this is pretty clear to see that we get all this, which is under the curve. But in addition, we also get this little portion outside. And if we add extra area with with every rectangle, that means that we're gonna overestimate the area. So therefore the midpoint overestimates. So that's the proof. And just summarizing. We know that the trap is oid underestimates. So it's less than the inner girl from A to B of after Lex DX. But we also know that the mid point over estimates. So this is exactly what we needed to show we proved it.

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