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∫_{0}^{1}1/√xdx = = lim_{a→0}∫_{a}^{1}1/√xdx Indefinite integral of f(x)=1/√x is F(x)=2√x. Indeed, let's take a derivative of F(x)=2√x: D_{x}F(x) = 2·(1/2)·x^{(1/2)−1} = = x^{−1/2} = 1/√x

Using Newton-Leibniz formula, ∫_{a}^{1}1/√xdx = F(1) − F(a) = = 2√1 − 2√a As a→0, this expression converges to 2

Answer: 2

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Example 1.2

∫_{0}^{∞}e^{−x}dx = = lim_{b→∞}∫_{0}^{b}e^{−x}dx Indefinite integral of f(x)=e^{−x} is F(x)=−e^{−x}. Evaluating integral: ∫_{0}^{b}e^{−x}dx = F(b) − F(0) = = (−e^{−b}) − (−e^{−0}) = 1 −e^{−b} As b→∞, this expression converges to 1

Answer: 1

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Example 1.3

∫_{0}^{∞}1/(1+x²) dx = = lim_{b→∞}∫_{0}^{b}1/(1+x²) dx Indefinite integral of f(x)=1/(1+x²) is F(x)=arctan(x). Evaluating integral: ∫_{0}^{b}1/(1+x²) dx = F(b) − F(0) = = arctan(b) − arctan(0) = = arctan(b) As b→∞, this expression converges to π/2

Answer: π/2

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Example 1.4

∫_{1}^{∞}(1/x²)·e^{1/x}dx = = lim_{b→∞}∫_{1}^{b}(1/x²)·e^{1/x}dx Indefinite integral of f(x)=(1/x²)·e^{1/x} can be found by noticing that −1/x² is a derivative of 1/x and, therefore, a derivative of e^{1/x} is e^{1/x}·(−1/x²). Therefore, indefinite integral of f(x)=(1/x²)·e^{1/x} is F(x)=−e^{1/x}. Evaluating integral: ∫_{1}^{b}(1/x²)·e^{1/x}dx = F(b)−F(1) = = (−e^{1/b}) − (−e^{1/1}) = e − e^{1/b} As b→∞, this expression converges to e −1

Recall the definition of the definite integral: ∫_{a}^{b}f(x) dx = = limΣ_{i∈[1,N]} f(x_{i})·Δx_{i} where Δx_{i}=x_{i}−x_{i−1} represents partitioning of segment [a,b] into N parts, and it is assumed that the widest interval Δx_{i} is shrinking to zero by length as N→∞.

Before we only defined and calculated these integrals for cases where the segment of definition [a,b] was finite and the integrated function f(x) was, at least, continuous on this segment and, as a consequence, was finite as well.

Consider now cases of integration over unbounded (infinite) domains and/or functions that go to infinity around some point(s) within their domains. Let's start with a statement that we never defined these types of integrals. Riemann sums are not applicable in these cases.

For example, you want to determine an area under the curve y=1/x² from left boundary x=1 to positive infinity. This function diminishes to zero as x→∞, so, intuitively, the area under this infinite curve might or might not be finite, depending on how fast the function value goes to zero as its argument goes to infinity.

Our first order is to define definite integrals of this kind.

Definition for infinite intervals of integration

We will define an improper integral of each kind as a limit of the corresponding proper integral if and only if this limit exists.

When both margins are infinite, we can define the integral as a sum of two integrals, each with only one margin being infinite, provided both exists in a sense of corresponding limits as defined above. 3. ∫^{∞}_{−∞} f(x) dx = = ∫^{0}_{−∞} f(x) dx + ∫^{∞}_{0} f(x) dx

Definition for functions going to infinity

Assume, we are integrating a function that asymptotically goes to infinity around one point within or on the border of a segment of integration. For example, ∫_{0}^{1}ln(x) dx As we know, logarithm goes to negative infinity as we approach argument 0, which is a left boundary of integration segment. To define this integral, we will cut off this point out of integration by stepping side-wise and take a limit of the result as the point of cut-off is getting closer and closer to a point where our function is not defined. If this special point is on the border of a segment of integration, we will have to take only one such limit. If it's in the middle, we will have to split the segment in two parts and integrate each one separately using this technique.

Assume, function f(x) is defined and continuous on interval (a,b] that is open on the left because f(x)→∞ as x→a. Then we define 4. ∫_{a}^{b}f(x) dx = = lim_{d→0}∫^{b}_{a+d} f(x) dx For any d, however small, integral on the right exists. So, if there is its limit as d→0, that limit is the definition of the integral on the left.

Analogously, assume, function f(x) is defined and continuous on interval [a,b) that is open on the right because f(x)→∞ as x→b. Then we define 5. ∫_{a}^{b}f(x) dx = = lim_{d→0}∫_{a}^{b−d}f(x) dx For any d, however small, integral on the right exists. So, if there is its limit as d→0, that limit is the definition of the integral on the left.

Finally, assume, function f(x) is defined and continuous on intervals [a,b) and (b,c] - everywhere at segment [a,c] except point x=b because f(x)→∞ as x→b. Then we define 6. ∫_{a}^{c}f(x) dx = = ∫_{a}^{b}f(x) dx + ∫_{b}^{c}f(x) dx provided both integrals on the right exist in a sense of limits defined above.

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Example 1

∫_{1}^{∞}1/x² dx = = lim_{b→∞}∫_{1}^{b}1/x² dx The indefinite integral (antiderivative) of f(x)=1/x² is function g(x)=−1/x. Therefore, the definite integral on the right in the above equality can be evaluated by Newton-Leibniz formula as ∫_{1}^{b}1/x² dx = = (−1/b) − (−1/1) = 1 − 1/b Now we can take a limit of this expression as b→∞: lim_{b→∞} (1 −1/b) = 1 Therefore, according to the definition of this improper integral, ∫_{1}^{∞}1/x² dx = 1 _______

Example 2

∫_{0}^{1}ln(x) dx = = lim_{d→0}∫_{d}^{1}ln(x) dx Indefinite integral (antiderivative) of f(x)=ln(x) is g(x)=x·ln(x)−x (refer to a lecture about indefinite integrals or just differentiate this expression to check that its derivative is indeed equal to ln(x)). Using Newton-Leibniz formula, we can evaluate the integral on the right: ∫_{d}^{1}ln(x) dx = = (1·ln(1)−1) − (d·ln(d)−d) = = −1 − d·ln(d) + d Going to a limit as d→0, we notice that lim_{d→0} d·ln(d) = 0 (see a note with the proof below) and, therefore, our integral, according to the definition, is ∫_{0}^{1}ln(x) dx = −1 NOTE: Proof of the limit: lim_{d→0} d·ln(d) = = lim_{d→0} ln(d)/(1/d) Use L'Hopital's rule to replace the ratio of functions with ratio of their derivatives. lim_{d→0} ln(d)/(1/d) = = lim_{d→0} (1/d)/(−1/d²) = = lim_{d→0} (−d) = 0

Definite Integrals - Examples of Integration Using Newton-Leibniz Formula

In this lecture we will consider exactly the same examples we used to demonstrate how definite integral can be calculated using its definition as a limit of sums.

Example 1

Find "area under curve" for f(x) = 10x on segment [a=0, b=4].

Solution

The indefinite integral (antiderivative) of function f(x) = 10x is 5x² (we can omit addition of a constant). This function should be evaluated at the end points of integration resulting in the following: ∫_{0}^{4}10x dx = = 5·4² − 5·0² = 80−0 = 80 This corresponds to the answer we have obtained previously using limit of Riemann sums. The end. ________________

Example 2

Find "area under curve" for f(x) = −x^{2}+1 on segment [a=−1, b=1].

Solution

The indefinite integral (antiderivative) of function f(x) = −x^{2}+1 is −x³/3+x (we can omit addition of a constant). This function should be evaluated at the end points of integration resulting in the following: ∫_{−1}^{1}(−x^{2}+1) dx = = [−(1)^{3}/3+(1)] − [−(−1)^{3}/3+(−1)] = = 2/3 − (−2/3) = 4/3 This corresponds to the answer we have obtained previously using limit of Riemann sums. The end. ________________

Example 3

The driver slows its car down by pressing the brakes from initial speed 20 meters per second to complete stop in 10 seconds, reducing its speed by the same value each second (linear dependency of the speed on time). Find the distance the car covered during this breaking process.

Solution

First of all, we have to find the formula that represents the speed as a function of time. Since every second the car slows down by the same number of meters per second, and it took 10 seconds to reduce the speed from 20 to 0, the function describing the speed is V(t)=20−2t. The indefinite integral (antiderivative) of function V(t)=20−2t is 20t−t² (we can omit addition of a constant). This function should be evaluated at the end points of integration [0,10] resulting in the following: ∫_{0}^{10}(20−2t) dx = = [20·10−10²] − [10·0−0²] = = 100 − 0 = 100 The distance cover by car during breaking is 100 meters. The end. ________________

Example 4

The Hooke's Law tells that a force needed to expand a string from its neutral position linearly depends on the length we expand it: F=K·x, where F is the force, x - the length of expansion and K - coefficient that depends on the physical properties of a spring. Given a spring with K=0.5 (in newtons per meter). Determine work W required to expand this spring by 0.1 meter.

Solution

Physical concept of work is defined as a product of a force by a distance this force is applied if this force is constant. In our case the force is changing with distance. To overcome this, we will approach this problem similarly to calculating an area under curve, where curve represents the force. Partition the distance (spring's expansion) into small intervals and assume that the force is constant on each interval. If the force is a function of distance F(x) then the amount of its work from distance x=0 to distance x=d can be represented as ∫_{0}^{d}F(x) dx

Applying this to our case, we have to calculate ∫_{0}^{0.1}0.5·x dx The indefinite integral of function 0.5·x is 0.25·x². Therefore, W = ∫_{0}^{0.1}0.5·x dx = = 0.25·0.1² − 0.25·0² = 0.0025 The work equals to 0.0025(joules) The end.

Recall the definition of the definite integral: ∫_{a}^{b}f(x) dx = = limΣ_{i∈[1,N]} f(x_{i})·Δx_{i} where Δx_{i}=x_{i}−x_{i−1} represents partitioning of segment [a,b] into N parts, and it is assumed that the widest interval Δx_{i} is shrinking to zero by length as N→∞.

First Fundamental Theorem of Calculus

Consider a smooth function f(x) on segment [a,b] and any point t∈[a,b]. The following definite integral can be considered as a function of t: F(t) = ∫_{a}^{t}f(x) dx The First Fundamental Theorem of Calculus states that the derivative of function F(t) is function f(t): F^{I}(t) = f(t)

Proof

Since, by definition, F^{I}(t) = = lim_{Δt→0}[F(t+Δt)−F(t)] /Δt we have to prove that lim_{Δt→0}[∫_{a}^{t+Δt}f(x) dx − − ∫_{a}^{t}f(x) dx] /Δt = f(t)

From the properties of definite integrals we know that ∫_{a}^{t}f(x) dx + ∫_{t}^{t+Δt}f(x) dx = = ∫_{a}^{t+Δt}f(x) dx from which follows that ∫_{a}^{t+Δt}f(x) dx − ∫_{a}^{t}f(x) dx = = ∫_{t}^{t+Δt}f(x) dx

Therefore, we have to prove that lim_{Δt→0}∫_{t}^{t+Δt}f(x) dx /Δt = f(t)

From properties of definite integrals we know that m·Δt ≤ ∫_{t}^{t+Δt}f(x) dx ≤ M·Δt where m is minimum of function f(x) on segment [t,t+Δt] and M - its maximum on this segment.

This allows us to state that the expression lim_{Δt→0}∫_{t}^{t+Δt}f(x) dx /Δt is bounded from below by m (minimum f(t) on segment [t,t+Δt]) and from above by M (maximum f(t) on segment [t,t+Δt]).

As Δt→0, our segment [t,t+Δt] shrinks to a point t. Since we assume sufficient "smoothness" of function f(t) (in this case we need just its continuity), both minimum and maximum of f(t) on segment [t,t+Δt] converge to the same value f(t). That forces the limit above also to converge to f(t).

End of proof.

IMPORTANT: Since the derivative of function F(t) above (definite integral of function f(x) on a segment from a to t) equals to f(t), function F(t) can serve as an indefinite integral (antiderivative) of f(t). Since we proved the existence and uniqueness of a definite (Riemann) integral for any continuous function, the theorem above has proven that for any continuous function there exists indefinite integral (its antiderivative).

Newton-Leibniz Formula

Let's assume that we want to find a definite (Riemann) integral ∫_{a}^{b}f(x) dx of some continuous function f(x) on segment [a,b]. Assume further that we know one particular function G(t) which is the indefinite integral (antiderivative) of f(t) (we deliberately decided to use different argument symbol t instead of x as an argument of function G(t) to have x only as a variable of integration). That is, G^{I}(t) = f(t).

Recall that there are many functions, derivative of which equal to f(t), but we know that all of them differ from each other only by an addition of a constant.

So, on one hand we have G(t) as one of the possible indefinite integrals (antiderivative) of f(t). On the other hand, we have just proven that derivative of F(t) = ∫_{a}^{t}f(x) dx by t also equals to function f(t). Therefore, we have two different functions, derivatives of both of which are the same function f(t): G^{I}(t) = f(t) and F^{I}(t) = [∫_{a}^{t}f(x) dx]^{I} = f(t)

Since two antiderivatives differ only by an addition of a constant, we conclude that F(t) = ∫_{a}^{t}f(x) dx = G(t) + C where C - some constant.

It's easy to find this constant. Since a definite integral on a null-segment [a,a] equals to zero, assign t=a in the above formula getting ∫_{a}^{a}f(x) dx = G(a) + C = 0 from which we get C = −G(a)

The final formula for our definite integral, therefore, is F(t) = ∫_{a}^{t}f(x) dx = G(t) − G(a) In particular, for t=b, we obtain the Newton-Leibniz formula: ∫_{a}^{b}f(x) dx = G(b) − G(a) where G(t) - any function, derivative of which is f(t), in other words, an indefinite integral (antiderivative) of f(t).

CONCLUSION

To find a definite integral of some real function f(t) on segment t∈[a,b], it is sufficient to find any particular indefinite integral (antiderivative) G(t) of function f(t) and calculate the expression G(b)−G(a). This establishes a connection between indefinite and definite integrals and justifies the usage of the same word integral for both.

Consider the following problem. A car moves along a straight line with variable speed given by function v(t) that defines speed v at any moment of time t. Our task is to find the distance covered by this car from time moment t=a to moment t=b.

If the speed is constant v(t)=V, the solution is easy: S = V·(b−a)

For variable speed the problem is not that easy. Here is what can be suggested as an approximate solution. Let's divide our time interval [a,b] into N equal short intervals [a=t_{0},t_{1}], [t_{1},t_{2}], [t_{2},t_{3}]... [t_{N−1},t_{N}=b] and assume that during each interval the speed is not significantly changing - a reasonable assumption if the time interval is small enough. Then the distance covered during i-th time interval [t_{i−1},t_{i}] is approximately equal to ΔS_{i} = v(t_{i})·(t_{i}−t_{i−1}) Here we use v(t_{i}) (the value on the right margin) as a constant speed during i-th time interval. We could have taken any other value during this interval - on the left margin, minimum on this interval, maximum or any in-between.

Notice that the approach is absolutely equivalent to our approach of finding the area under curve in the previous lecture.

Our next step is to summarize all ΔS_{i} to get a total distance and go to a limit as the number of intervals we divide our total travel time increases to infinity. So, the final formula is S = limΣ_{i∈[1,N]} v(t_{i})·(t_{i}−t_{i−1}) = = limΣ_{i∈[1,N]} v(t_{i})·Δt_{i} where limit is assumed to be taken when N→∞ and maximum width among all intervals Δt_{i} diminishes to zero.

As with a problem of area under curve, we can prove that the limit is independent of which point within each interval is used to get the speed value. This limit is also independent on how we break the total time travel into shorter intervals as long as the length of the longest among intervals is shrinking to zero as we proceed with dividing the total travel time into more and more intervals.

Draining

Consider the following problem. There is a tub filled with water. When we open the drain, the water starts flowing out of the tub. The speed of water flow depends on different factors - some constant (the size of a drain pipe) and some variable (the pressure at the drain opening).

Our task is to determine the volume of water that is drained from the tub during some known period of time from t=a to t=b, provided we know the speed of draining v(t) (in some units, like liters per second) at any moment of time t.

The complication here, obviously, is that the speed of water flow through a drain is variable because it depends on the water pressure at the drain opening, which changes as the water flows out of the tub.

Our approach to this problem is similar to the one above. Let's divide our time interval [a,b] into N equal short intervals [a=t_{0},t_{1}], [t_{1},t_{2}], [t_{2},t_{3}]... [t_{N−1},t_{N}=b] and assume that during each interval the speed of water flow through a drain is not significantly changing - a reasonable assumption if the time interval is small enough. Then the volume of water going through a drain during i-th time interval [t_{i−1},t_{i}] is approximately equal to ΔW_{i} = v(t_{i})·(t_{i}−t_{i−1}) Here we use v(t_{i}) (the value on the right margin) as a constant speed during i-th time interval. We could have taken any other value during this interval - on the left margin, minimum on this interval, maximum or any in-between.

Notice that the approach is absolutely equivalent to our approach of finding the area under curve in the previous lecture.

Our next step is to summarize all ΔW_{i} to get a total volume of drained water and go to a limit as the number of intervals we divide our total drainage time increases to infinity. So, the final formula is W = limΣ_{i∈[1,N]} v(t_{i})·(t_{i}−t_{i−1}) = = limΣ_{i∈[1,N]} v(t_{i})·Δt_{i} where limit is assumed to be taken when N→∞ and maximum width among all intervals Δt_{i} diminishes to zero.

As with a problem of area under curve, we can prove that this limit is independent of which point within each interval is used to get the speed value. This limit is also independent on how we break the total time of drainage into shorter intervals as long as the length of the longest among intervals is shrinking to zero as we proceed with dividing the total time into more and more intervals.

Volume of Solids of Revolution

Consider the following problem. There is a solid obtained by a revolution of some curve on a plane around an axis which also lies on this plane. Let's assume that the curve is defined as a graph of a smooth function y=f(x), where argument x varies from a to b, and the axis of rotation is the X-axis.

Our task is to determine the volume of this solid.

If our function is constant, that is f(x)=c, our solid is a cylinder of height H=b−a and radius equal to that constant c. If function f(x) is linear (so, its graph is a straight line), our solid is a truncated cone of height H=b−a and radiuses of its two bases equal to R_{a}=f(a) and R_{b}=f(b). In either of the above cases we know classic geometric formulas to calculate the volume of a solid. The complexity of our problem, obviously, is that the shape of our solid is not one of those well known types.

Let's approach our problem analogously to determining the area under curve. Let's divide our argument interval [a,b] into N equal short intervals [a=x_{0},x_{1}], [x_{1},x_{2}], [x_{2},x_{3}]... [x_{N−1},x_{N}=b] and assume that on each interval the value of function f(x) is not significantly changing - a reasonable assumption if the interval is small enough.

Replacing the function values on each interval [x_{i−1},x_{i}] with a constant value at its right margin f(x_{i}) and rotating the obtained step-function around the X-axis, we obtain a different solid, but the one that approximates the original one to certain degree. The approximation will be better if the number of intervals we divide segment [a,b] is increasing and the size of the largest interval diminishes to zero.

The result of rotation of a step-function on each interval will be a cylinder with height h_{i}=x_{i}−x_{i−1} and radius f(x_{i}), its volume will be equal to ΔV_{i} = π·f ^{2}(x_{i})·(x_{i}−x_{i−1}) = = π·f ^{2}(x_{i})·Δx_{i}

Our next step is to summarize all ΔV_{i} to get a total volume of a solid and go to a limit as the number of intervals we divide our segment [a,b]. So, the final formula is V = limΣ_{i∈[1,N]} π·f ^{2}(x_{i})·Δx_{i} where limit is assumed to be taken when N→∞ and maximum width among all intervals Δx_{i} diminishes to zero.

As with a problem of area under curve, we can prove that this limit is independent of which point within each interval is used to get the radius of a cylinder. This limit is also independent on how we break the total range of arguments into shorter intervals as long as the length of the longest among intervals is shrinking to zero as we proceed with dividing the total range into more and more intervals.

CONCLUSION

In all the cases we considered so far we came up with an expression limΣ_{i∈[1,N]} f(x_{i})·Δx_{i} where f(x) is some smooth function defined on a segment [a,b], {x_{i}} is partitioning of segment [a,b] into N parts, and we assume that N→∞ and the maximum width of intervals Δx_{i}=x_{i}−x_{i−1} converges to zero.

We have also proven that this limit exists for any smooth function f(x), that it does not depend on how we partition our segment [a,b] (as long as the widest interval of division shrinks to zero length) and it does not depend on which point within each interval of division we use to determine the function value on this interval.

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