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Question 3: Evaluate ∫ 0 1 (xe x + sin (πx/4) dxĪnswer : Let I = ∫ 0 1 (xe x + sin (πx/4) dx. Using the standard integral ∫ dx / (√) = sin –1 (x/a), we get Question 1: Evaluate ∫ 0 1 dx / (√)Īnswer : Let I = ∫ 0 1 dx / (√). To solve the indefinite integral, let’s substitute sin2t = u, so that 2 cos2t dt = du or cos2t dt = ½ du. Solution: Let I = ∫ 0 π/4 sin 32t cos2t dt Using the rules of partial fractions, we get Solution: Let I = ∫ 1 2 x dx / (x + 1)(x + 2) Hence, using the second fundamental theorem of calculus, we get
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Let’s substitute (30 – x 3/2) = t, so that – 3/2 √x dx = dt or √x dx = – 2/3 dt. Solution: Let I = ∫ 4 9 dxįirst, we find the anti-derivative of the integrand. Using the second fundamental theorem of calculus, we get
![fundamental theorem of calculus part 2 fundamental theorem of calculus part 2](https://d2vlcm61l7u1fs.cloudfront.net/media/968/96817391-8c49-4c1b-a219-5333fdd821b2/phpx50v2l.png)
Now, the indefinite integral, ∫ x 2 dx = x 3/3 = F(x) That’s it! Let’s look at some examples now. There is no need to keep the integration constant C because it disappears while evaluating the value of the definite integral.
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In ∫ a b f(x) dx, the function ‘f’ should be well defined and continuous in.It strengthens the relationship between differentiation and integration. The most important step in evaluating a definite integral is finding a function whose derivative is equal to the integrand.This theorem is useful because we can calculate the definite integral without calculating the limit of a sum.In words, the Theorem 2 tells us that ∫ a b f(x) dx = (value of the anti-derivative ‘F’ of ‘f’ at the upper limit b) – (value of the same anti-derivative at the lower limit a).∫ a b f(x) dx = a b = F(b) – F(a) Important Points to Remember If ‘f’ is a continuous function defined on the closed interval and F is an anti-derivative of ‘f’. If ‘f’ is a continuous function on the closed interval and A (x) is the area function. This helps us define the two basic fundamental theorems of calculus. This is denoted by A(x) and represented as follows: The area of this shaded region depends on the value of ‘x’. However, the assertions made below are true for other functions as well. Note: We are assuming that f(x)> 0 for x ∈. If ‘x’ is a point in, then ∫ a x f(x) dx represents the area of the shaded region in the figure above. In this case, t is simply a placeholder that allows us to avoid confusion that could otherwise result from using x in both the bounds of the integral as well as the integrand since for all values of x, F(x) is dependent only on x, not t.By definition, ∫ a b f(x) dx is the area of the region bounded by the curve y = f(x), the x-axis and the coordinates ‘x = a’ and ‘x = b’. Note that even though F(x) is defined in terms of the integral of another function, which may seem confusing since we expect a definite integral to be a number, the result of the integral for any given x value is indeed a number. In other words, the area under the curve is represented by: In the figure, F(x) is a function that represents the area under the curve between a and some point x within the interval. Given a function f(t) that is continuous over an interval, recall that an integral represents the area under the curve. It also gives us an efficient way to evaluate definite integrals. The fundamental theorem of calculus (FTC) establishes the connection between derivatives and integrals, two of the main concepts in calculus. Home / calculus / integral / fundamental theorem of calculus Fundamental theorem of calculus