![]() x exp(- x) - exp(- x) + C The Fundamental Theorem of Calculus Let u = x, dv = exp(- x), therefore du = dx, v = -exp(- x) Given some product to integrate, we arrange for U and dV to make the integral on the right-hand side, V( dU/ dx) more simple than the integral U ( dV/ dx) we started with. Integrating both sides with respect to x and rearranging, We must change the limits of integration, the new values come from u = x 3, therefore when x= 1, u = 1 and when x= 2, u = 8. The substitution of a function, may simplify the integral allowing it to be calculated easily. ∫cosec( u) du = ln|cosec( u) + cot( u)| + C Therefore, when we integrate, we have to add a constant because differentiation of a constant is zero.The value of the constant has to be determined by additional information about the equation, for example where it intercepts the y-axis. We would also obtain the same answer for x 3/3 + 5 or x 3/3 - 2 or x 3/3 + any constant. Why is this? If we take our answer x 3/3 and differentiate with respect to x, we obtain x 2. The integration is performed in the same way but we must remember to add an arbitrary constant known as the constant of integration. Rather, the result is a family of functions. Where there are no limits on the integral sign, the integral is called indefinite, meaning there is no specific value. The final result is given by taking the first limit from the second. Next the lower limit, 2, is substituted for x giving, 2 3/3 = 8/3. The value of the top limit is substituted in for x in the result. These are called the the limits of integration, the top one is known as the uppper limit and the bottom one is the lower limit.Įssentially, these numbers are substituted into the integral after the integration has been performed. The integration symbol sometimes has numbers or other letters alongside the integral sign. An overestimate is cancelled out to some degree by the underestimate.ĭefinite and Indefinite Integrals Definite Integrals The accuracy for larger rectangles widths can be improved by choosing the mid-point of each rectangle, to corespond to the value of the function at that point. In the limit, the sum becomes the definite integral of f( x). The areas calculated by the left and right Riemann sums become closer to the true value of the area. In the limit, Δ x tends to zero as n tends to ∞. Approximating the area under a curve by rectangular elements.Īs the number of strips increase, the value of Δ x decreases. Right-Riemann SumĪ r = f( x 1)Δ x + f( x 2)Δ x +. ![]() In this case, it over-estimates the area under the curve. We could also have represented the area by starting our summation of rectangles at k=1 and endding at k= n. Left-Riemann SumĪ l = f( x 0)Δ x + f( x 1)Δ x +. In this case resulting area will be an under-estimate of the true area. The left-Riemann sum starts with the zeroth strip and ends at the ( n-1)th strip. Summing all the areas of all the strips gives the approximate area under the curve. ![]() Having partitioned the interval into n equal parts, the area of the kth strip is Δ x f(x k), where f( x k) is the value of the function at the x k. We define an interval which is divided into n equal parts. Imagine trying to find the area under a curve.Ĭlearly, this is a very poor approximation but we could do a little better by adding together the area of two rectangles of half the widths with different heights corresponding to the value of the function at the points where corner meets the function. ![]() One way of understanding integration is to see how it is used to find the area under a curve defined by a function f( x). In fact you can't really do any serious physics without knowing about it. Integration is widely used throughout mathematics and physics and so is an important concept to grasp. In fact the integration symbol derrivesfrom an elongated letter S, first used by Leibniz, to stand for Summa meaning sum in Latin. It is an extension of the concept of summation. Integration is a major parts of calculus. ![]()
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