The connection between the definite integral and indefinite integral is given by the second part of the Fundamental Theorem of Calculus. Instead the function is not continuous because it takes on different values on either sides of \(x = 1\). Recall from our first example above that all we really need here is any anti-derivative of the integrand. Now, in the first integrals we have \(t < \frac{5}{3}\) and so \(3t - 5 < 0\) in this interval of integration. Often times they won’t. On each of these intervals the function is continuous. We just computed the most general anti-derivative in the first part so we can use that if we want to. In the second term, taking the 3 out of the denominator will just make integrating that term easier. First, notice that we will have a division by zero issue at \(w = 0\), but since this isn’t in the interval of integration we won’t have to worry about it. This is here only to make sure that we understand the difference between an indefinite and a definite integral. Note that this problem will not prevent us from doing the integral in (b) since \(y = 0\) is not in the interval of integration. If even one term in the integral can’t be integrated then the whole integral can’t be done. For this integral notice that \(x = 1\) is not in the interval of integration and so that is something that we’ll not need to worry about in this part. In fact we can say more. Note that the absolute value bars on the logarithm are required here. This was also a requirement in the definition of the definite integral. We didn’t make a big deal about this in the last section. The constant that we tacked onto the second anti-derivative canceled in the evaluation step. Neither of these are terribly difficult integrals, but we can use the facts on them anyway. Surprisingly, these questions are related to the derivative, and in some sense, the answer to each one is the opposite of the derivative. The following is a table of formulas of the commonly used Indefinite Integrals. Evaluate each of the following integrals. Compare this answer to the previous answer, especially the evaluation at zero. The question is asking "what is the integral of x3 ?". This means that the integrand is no longer continuous in the interval of integration and that is a show stopper as far we’re concerned. This is the last topic that we need to discuss in this section. You appear to be on a device with a "narrow" screen width (, Derivatives of Exponential and Logarithm Functions, L'Hospital's Rule and Indeterminate Forms, Substitution Rule for Indefinite Integrals, Volumes of Solids of Revolution / Method of Rings, Volumes of Solids of Revolution/Method of Cylinders, Parametric Equations and Polar Coordinates, Gradient Vector, Tangent Planes and Normal Lines, Triple Integrals in Cylindrical Coordinates, Triple Integrals in Spherical Coordinates, Linear Homogeneous Differential Equations, Periodic Functions & Orthogonal Functions, Heat Equation with Non-Zero Temperature Boundaries, Absolute Value Equations and Inequalities, \( \displaystyle \int{{{y^2} + {y^{ - 2}}\,dy}}\), \( \displaystyle \int_{{\,1}}^{{\,2}}{{{y^2} + {y^{ - 2}}\,dy}}\), \( \displaystyle \int_{{\, - 1}}^{{\,2}}{{{y^2} + {y^{ - 2}}\,dy}}\), \( \displaystyle \int_{{ - 3}}^{1}{{6{x^2} - 5x + 2\,dx}}\), \( \displaystyle \int_{{\,4}}^{{\,0}}{{\sqrt t \left( {t - 2} \right)\,dt}}\), \( \displaystyle \int_{{\,1}}^{{\,2}}{{\frac{{2{w^5} - w + 3}}{{{w^2}}}\,dw}}\), \( \displaystyle \int_{{\,25}}^{{\, - 10}}{{dR}}\), \( \displaystyle \int_{{\,0}}^{{\,1}}{{4x - 6\sqrt[3]{{{x^2}}}\,dx}}\), \( \displaystyle \int_{{\,0}}^{{\,\frac{\pi }{3}}}{{2\sin \theta - 5\cos \theta \,d\theta }}\), \( \displaystyle \int_{{\,{\pi }/{6}\;}}^{{\,{\pi }/{4}\;}}{{5 - 2\sec z\tan z\,dz}}\), \( \displaystyle \int_{{\, - 20}}^{{\, - 1}}{{\frac{3}{{{{\bf{e}}^{ - z}}}} - \frac{1}{{3z}}\,dz}}\), \( \displaystyle \int_{{\, - 2}}^{{\,3}}{{5{t^6} - 10t + \frac{1}{t}\;dt}}\), \(\displaystyle \int_{{\,10}}^{{\,22}}{{f\left( x \right)\,dx}}\), \(\displaystyle \int_{{\, - 2}}^{{\,3}}{{f\left( x \right)\,dx}}\), \( \displaystyle \int_{{\, - 2}}^{{\,2}}{{4{x^4} - {x^2} + 1\,dx}}\), \( \displaystyle \int_{{\, - 10}}^{{\,10}}{{{x^5} + \sin \left( x \right)\,dx}}\). The connection between the definite integral and indefinite integral is given by the second part of the Fundamental Theorem of Calculus. Much easier than using the definition wasn’t it? Therefore, the desired function is f(x)=1 4 Embedded content, if any, are copyrights of their respective owners. In order to do this one will need to rewrite both of the terms in the integral a little as follows. In this section we will take a look at the second part of the Fundamental Theorem of Calculus. Let’s work a couple of examples that involve other functions. Note that the limits of integration are important here. Please submit your feedback or enquiries via our Feedback page. If we know the f’ of a function which is differentiable in its domain, we can then calculate f. In differential calculus, we used to call f’, the derivative of the function f. Here, in integral calculus, we call f as the anti-derivative or primitive of the function f’.
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