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## Application of Definite Integrals

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**Application of Definite Integrals**Dr. Farhana Shaheen Assistant Professor YUC- Women Campus**Calculus (Latin, calculus, a small stone used for counting)**• Calculus is a branch of mathematics with applications in just about all areas of science, including physics, chemistry, biology, sociology and economics. Calculus was invented in the 17th century independently by two of the greatest mathematicians who ever lived, the English physicist and mathematician Sir Isaac Newton and the German mathematician Gottfried Leibniz. • Calculus allows us to perform calculations that would be practically impossible without it.**Calculus is the study of change**• Calculus is a discipline in mathematics focused on limits, functions, derivatives, integrals, and infinite series. Calculus is the study of change, in the same way that geometry is the study of shape and algebra is the study of operations and their application to solving equations. • This subject constitutes a major part of modern mathematics education. It has widespread applications in science, economics, and engineering and can solve many problems for which algebra alone is insufficient.**Calculus is a very versatile and valuable tool. It is a form**of mathematics which was developed from algebra and geometry. It is made up of two interconnected topics: • i) Differential calculus • ii) Integral calculus.**Differential calculus is the mathematics of motion and**change. • Integral calculus covers the accumulation of quantities, such as areas under a curve or volumes between two curves. • Integrals and derivatives are the basic tools of calculus, with numerous applications in science and engineering. The two ideas work inversely together in Calculus. • We will discuss about integration and its applications.**INTEGRATION**• Integration is an important concept in Mathematics and, together with differentiation, is one of the two main operations in Calculus. • A rigorous mathematical definition of the integral was given by Bernhard Riemann.**Definite and Indefinite Integrals**• Integration may be introduced as a means of finding areas using summation and limits. This process gives rise to the definite integral of a function. • Integration may also be regarded as the reverse of differentiation, so a table of derivatives can be read backwards as a table of anti-derivatives. The final result for an indefinite integral must, however, include an arbitrary constant, because there is a family of curves having same derivatives, i.e. same slope.**Indefinite Integrals**• The term Indefinite integral is referred to the notion of antiderivative, a function F whose derivative is the given function ƒ. In this case it is called an indefinite integral. Some authors maintain a distinction between anti-derivatives and indefinite integrals.**Indefinite Integral as Net Change**• Problem 1:A particle moves along the x-axis so that its acceleration at any time t is given by a(t) = 6t - 18. At time t = 0 the velocity of the particle is v(0) = 24, and at time t = 1 its position is x(1) = 20. (a) Write an expression for the velocity v(t) of the particle at any time t. (b) For what values of t is the particle at rest? (c) Write an expression for the position x(t) of the particle at any time t. (d) Find the total distance traveled by the particle from t = 1 to t = 3.**Solution:**• (a) v(t) = ∫ a(t) dt = ∫ (6t - 18) dt = 3t2 - 18t + C 24 = 3(0)2 - 18(0) + C 24 = C so v(t) = 3t2 - 18t + 24 • (b) The particle is at rest when v(t) = 0. 3t2 - 18t + 24 = 0 t2 - 6t + 8 = 0 (t - 4)(t - 2) = 0 t = 4, 2(c) x(t) = ∫ v(t) dt = ∫ (3t2 - 18t + 24) dt = t3 - 9t2 + 24t + C 20 = 13 - 9(1)2 + 24(1) + C 20 = 1 - 9 + 24 + C 20 = 16 + C 4 = Cso x(t) = t3 - 9t2 + 24t + 4**Calculating the Area of Any Shape**• Although we do have standard methods to calculate the area of some known shapes, like squares, rectangles, and circles, but Calculus allows us to do much more. Trying to find the area of shapes like this would be very difficult if it wasn’t for calculus.**Riemann integral**• The Riemann integral is defined in terms of Riemann sums of functions with respect to tagged partitions of an interval. Let [a,b] be a closed interval of the real line; then a tagged partition of [a,b] is a finite sequence • This partitions the interval [a,b] into n sub-intervals [xi−1, xi] indexed by i, each of which is "tagged" with a distinguished point tiε [xi−1, xi]. A Riemann sum of a function f with respect to such a tagged partition is defined as**Riemann integral**• Approximations to integral of √x from 0 to 1, with ■ 5 right samples (above) and ■ 12 left samples (below)**Riemann sums**• Riemann sums converging as intervals halve, whether sampled at ■ right, ■ minimum, ■ maximum, or ■ left.**Definite integral**• Given a function ƒ of a real variable x and an interval [a, b] of the real line, the definite integral is defined informally to be the net signed area of the region in the xy-plane bounded by the graph of ƒ, the x-axis, and the vertical lines x = a and x = b.**Measuring the area under a curve**• Definite Integration can be thought of as measuring the area under a curve, defined by f(x), between two points (here a and b).**Definite integral of a function**• A definite integral of a function can be represented as the signed area of the region bounded by its graph.**Definite integral of a function**• The principles of integration were formulated independently by Isaac Newton and Gottfried Leibniz in the late 17th century. Through the fundamental theorem of calculus, which they independently developed, integration is connected with differentiation as:**Fundamental Theorem of Calculus**• Let f(x) be a continuous function in the given interval [a, b], and F is any anti-derivative of f on [a, b], then**Area between two curves y = f(x) and y = g(x)**DEFINITION • If f and g are continuous and f (x) ≥ g(x) for a ≤ x ≤ b, then the area of the region R between f(x) and g(x) from a to b is defined as**Definite Integrals to find the Volumes**• We can also use definite integrals to find the volumes of regions obtained by rotating an area about the x or y axis.**Solid of Revolution**• A solid that is obtained by rotating a plane figure in space about an axis coplanar to the figure. The axis may not intersect the figure. • Example: Region bounded between y = 0, y = sin(x), x = π/2,x = π.**Volumes by slicing**I- Disk Method • A technique for finding the volume of a solid of revolution. This method is a specific case of volume by parallel cross-sections. II- Washer Method • Another technique used to finding the volume of a solid of revolution. The washer method is a generalized version of the disk method. • Both the washer and disk methods are specific cases of volume by parallel cross-sections.**Volumes by Disk method**• Let S be a solid bounded by two parallel planes perpendicular to the x-axis at x = a and x = b. If, for each x in [a, b], the cross- sectional area of S perpendicular to the x-axis is A(x) = π(f(x))2 ,then the volume of the solid is**Example: Right circular cone**Find the volume of the region boundedbetween y = 0,x = 0, y = -2x+3. Here r = 3. So, f(x) = -2x + 3 in the interval [0, 3]**For f(x) = 2 + Sin x, revolved about the x – axis**The volume is**Volumes by washer method perpendicular to the x-axis**• The volume of the solid generated when the region R, (bounded above by y=f(x) and below by y=g(x)), is revolved about the x-axis, is given by**Volumes by washer method perpendicular to the y-axis**• The volume of the solid generated when the region R (bounded above by x=f(y) and below by x=g(y)), is revolved about the y-axis, is given by**Figure illustrates how a washer can be generated from a**disk. We begin with a disk with radius rout and thickness h. A smaller concentric disk with radius rin is removed from the original disk. The resulting solid is a washer.**The Washer Method for Solids of RevolutionVolume of region**bounded between , y = x2.**When the solid is formed by revolving the region between the**graphs of y = f(x) and y = g(x), where f(x) > g(x), about the y-axis, the height of the rectangle is given by h = f(x)-g(x).