Integration Area Under a Curve - Assignment Example

Integration Area under a Curve Introduction Throughout this research project, my key aim would be to explore the subject of integration and its context in a manner intended to enlighten students and like-minded persons on the modalities of addressing such questions when confronted with any in the future. For a presentation in the class and in relation to determining a suitable teaching aspect, I will seek to develop lesson plans that will be presented in the form of PowerPoint expositions to the class. Subsequently, I will produce worksheets for the activities carried out and provide work-based practice exercises that will aid in strengthening the knowledge power of the students in relation to integration concepts. I will expand on the previous concepts presented before the class and discuss the relational recurrences noted. Integration Integration, as a mathematical phenomenon has been widely used in the calculation of areas as under a curve in a plane (graph); hence, it is an important concept applied in mathematical computations. In a simple case, the area under the curve is usually given by a single integral that is definite. However, in cases where the integral provides negative answers or in cases that are more complicated, the correct answer is usually obtained through the splitting of the area under the curve into several parts by the addition or subtraction of appropriate integrals (Berresford & Andrew 440-447). Therefore, this shows that for one to master these techniques, undertaking of plenty of practice exercises will be quite vital. After reading through this guideline report, one should be able to find: a. The area between the x-axis, the given two coordinates and the curve. b. The area existing between two curves. c. The area between the x-axis and the curve given that the ordinate points are those in which the axis is crossed by the curve. The area between the x-axis and the curve We shall begin by exploring the following equation: y = 6x5 + 6. To find the area under this curve between the value x = -1 and x = 2, we can follow the following steps. Integrate the given function. f(x) = 6x5 + 6 → F(x) = x6 + 6x Substitute the limits one by one. This is based on the notation given below. Subtract the answers from above. The values are x = -1 and x = 2. Substituting these by the values a and b, a = -1 and b = 2. This implies that the area can be got by, = [x6 + 6x]2 – [x6 + 6x]-1 = [64 + 12] – [-1 – 6] = 76 – (-7) = 83 Graphically, this area can be represented as below. From the above graph, the curve is shown as crossing the x-axis only once at the point where x = -1 and the y-axis at point y = 6. From the graph, it is easy to note that this required area bounded by the curve and the two points is the shaded region. This is the case of all the parts being above the x-axis. When asked to work out the area under a curve, it is essential that one remembers the notation shown above. In determining the area under the curve, we can perform the definite integral or an indefinite integral between the two points as shall be shown later on in this report (Das 220-228). Another calculation for the area under the curve can be presented as below. y = 20 – 3x2 x = 1 and x = 2 f(x) = 20 – 3x2 F(20x – x3) Presenting the integrals as a = 1 and b = 2, The area = [20x – x3]2 – [20x – x3]1 = (40 - 8) – (20 - 1) = 13 square units. In another example, the area under the curve may be presented as follows for the x-axis and the lines x = 1 and 5 for lower and upper limits respectively and the curve of y = x2 + 1 as shown below by the shaded region. Given this graph for the curve y = x2 + 1, it is possible to consider the area under the curve by sub-dividing the shaded region into four rectangles of equal width as shown below. As such, the estimated value of the area would take into consideration the area of the rectangles and estimate the same as 2+ 5 + 10 + 17 = 34 units. However, this value is an underestimate; thus, finding an overestimate value would entail taking the values as presented in the graph below: Thus, the corresponding estimate of the area becomes: A = 5 + 10 + 17 + 26 = 58 square units. Taking both estimates, equality for A (Area) can be formed as follows: 34 < A < 58. This shows that A satisfies this equality. From this, it is indicative that the exact value for the area lies within the formed inequality. Performing this operation using rectangles would require forming more rectangles of smaller sizes, which would have been tiresome. A better way would be through the general formulation of this problem, so as to provide an exactness (Berresford & Andrew 440-447). General formulation of integration Considering the example in the above case, this would generally be presented as: Definite integrals A definite integral has both upper and lower limits in the form presented below. Considering these two scenarios, it is evident that an integration based on definite integrals is guided by certain principles. First, and that For c ϵ (a, b) given that c is a constant. These principles need to be ensured in the answering of definite integration problems. Indefinite integrals With the case of indefinite integrals, the integral symbol can be used without necessarily applying the limits in the denoting that integration is being performed (Berresford & Andrew 440-447). For instance, earlier on in this study, it was shown that dy/dx = 2x → y = x2 + c. In this, an alternative way by which it can be expressed is: ∫2x dx = x2 + c. This is usually read as ‘the integral of 2x with respect to x. This presentation is the case of an indefinite integral computational exercise. Area below the x-axis Certain times, the curve may go below the x-axis. The outright expectation in this would be the curve forming a U-shaped with the x-axis. In this case, when the curve goes below the x-axis, the value of the corresponding y-axis becomes negative hence, resulting in a negative value of y δx (Berresford & Andrew 440-447). Therefore, the area of concern also includes this area under the x-axis. When performing integral calculations, if the integral becomes negative, the likely conclusion would be that the area is beneath the x-axis as presented in the graphs below. Works Cited Berresford, Geoffrey C, and Andrew M. Rockett. Brief Applied Calculus. Pacific Grove, Calif: Brooks/Cole/Cengage Learning, 2013. Print. Das, Shantanu. Functional Fractional Calculus. Berlin: Springer-Verlag Berlin, 2011. Print. Read More