Question

Let \(\mathrm{F}(\mathrm{x})=\int_{0}^{\mathrm{x}} \mathrm{f}(\mathrm{t}) \mathrm{dt}\). Determine a formula for computing \(\mathrm{F}(\mathrm{x})\) for all real \(\mathrm{x}\) if \(\mathrm{f}\) is defined as follows: (a) \(\mathrm{f}(\mathrm{t})=(\mathrm{t}+\mid \mathrm{t})^{2}\) (b) \(f(t)=\left\\{\begin{array}{lll}1-t^{2} & \text { if } & |t| \leq 1 \\\ 1-|t| & \text { if } & |t|>1\end{array}\right.\) (c) \(\mathrm{f}(\mathrm{t})=\mathrm{e}^{-1}\). (d) \(\mathrm{f}(\mathrm{t})=\) the maximum of 1 and \(\mathrm{t}^{2}\).

Short answer

In summary, the expressions for F(x) for each case are: (a) F(x) = (4/3)x³ for x≥0, and 0 for x<0 (b) F(x) = x - (1/3)x³ if |x| ≤ 1, 2/3 + x - x²/2 if x>1, and -2/3 - x - x²/2 if x<-1 (c) F(x) = xe⁻¹ (d) F(x) = x for x² ≤ 1, and (1/3)x³ for x² > 1

Step-by-step solution

Breaking down the function into two cases for positive and negative t

Since |t| takes the absolute value of t, we have to split the integration into two cases, one for t≥0 and another for t<0: For t≥0, |t| = t, and f(t) = (t + t)² = (2t)² = 4t² For t<0, |t| = -t, and f(t) = (t-t)² = 0 Now, we can proceed with the integration.

Integrate the function for each case

For t≥0, F(x) = ∫₀^x 4t² dt For t<0, F(x) = ∫₀^x 0 dt

Find the formula for F(x)

For t≥0, F(x) = (4/3)x³ + C For t<0, F(x) = 0 So, F(x) = (4/3)x³ for x≥0, and 0 for x<0. #Case (b): f(t) = 1 - t² if |t| ≤ 1, and 1 - |t| if |t| > 1#

Integrate the function for each case

For |t|≤1, F(x) = ∫₀^x (1 - t²) dt For |t|>1, F(x) = ∫₀^x (1 - |t|) dt For this case, we split the integration into two more cases: When x>1, and x<-1.

Find formulas for F(x) for each situation

For |t|≤1, F(x) = [t - (1/3)t³]₀^x = x - (1/3)x³ When x>1, we integrate from 0 to x for the function 1-t², and separately from x to 1 for the function 1-t, and add the results: F(x) = ∫₀^1 (1 - t²) dt + ∫₁^x (1 - t) dt = [t - (1/3)t³]₀^1 + [-t²/2 + t]₁^x = 2/3 + x - x²/2 When x<-1, similarly: F(x) = ∫₀^1 (1 - t²) dt + ∫₁^{|x|} (t - 1) dt = [t - (1/3)t³]₀^1 + [t²/2 - t]₁^{|x|} = -2/3 - x - x²/2 So, we have three cases for F(x): F(x) = x - (1/3)x³ if |x| ≤ 1, 2/3 + x - x²/2 if x>1, and -2/3 - x - x²/2 if x<-1. #Case (c): f(t) = e⁻¹#

Integrate the function

F(x) = ∫₀^x e⁻¹ dt

Find the formula for F(x)

Since e⁻¹ is a constant, F(x) = e⁻¹(x-₀) = xe⁻¹. #Case (d): f(t) = max{1, t²}#

Integrate the function for each case

For 1: F(x) = ∫₀^x 1 dt For t²: F(x) = ∫₀^x t² dt

Find the formula for F(x) for each case

For 1: F(x) = x For t²: F(x) = (1/3)x³ Now, the maximum of the two functions is 1 for x²≤1 and t² for x²>1. So, we have two cases for F(x): F(x) = x for x² ≤ 1, and (1/3)x³ for x² > 1.

Key concepts

Definite Integration

Definite integration represents the accumulation of quantities, like area under a curve, between two definite points. When we write \begin{align*}F(x) = \int_{0}^{x} f(t) dt,\end{align*}we're looking for the total accumulation of the function \(f(t)\) from the lower limit of integration, which is 0, to the upper limit, which is \(x\). For students tackling IIT JEE problems, understanding the geometric interpretation of definite integration is crucial. It's the foundational block for solving more complex problems, including those with piecewise functions or unique properties of integrals.

When practicing for the IIT JEE, be mindful of the limits of integration and what they signify in the context of the problem. Often, changing the limits can significantly alter the result. If the integrand, the function being integrated, is given by different rules over different intervals, you need to break down the integral and calculate it over each interval separately, then sum up the results.

In the provided example, the function \(f(t)\) changes depending on the value of \(t\) and requires careful consideration of the range over which we are integrating. The exercise also exemplifies the fundamental theorem of calculus, which connects the concept of differentiation and integration, a pivotal concept in IIT JEE syllabus.

Piecewise Functions Integration

Piecewise functions have different expressions for different intervals of the variable. They can appear daunting at first, but integrating them is a matter of handling each piece separately, according to its interval. The IIT JEE syllabus often includes Integration of piecewise functions because they test a student's ability to understand and apply integration techniques over different intervals.

Here's a tip for students: when faced with a piecewise function, sketch it out. It’s a simple way to visualize the different parts of the function. After breaking the range of integration into appropriate intervals, integrate each segment over its respective interval and combine the results, being careful not to miss any intervals.

In the given exercise, \(f(t)\) changes form based on the range of \(t\). For instance, in part (b), it is stipulated that the form of the function switches based on whether \(|t|\) is less than or equal to 1, or greater than 1. Therefore, the integration is executed in parts and the results from different intervals are appended to find the total integral for any real \(x\). This shows how handling the piecewise nature of functions is essential to master for definite integration in IIT JEE exams.

Properties of Definite Integrals

Properties of definite integrals streamline the process of integration and often allow us to evaluate integrals more quickly and easily. Understanding and harnessing these properties is a vital skill for any student aiming to excel in the IIT JEE. Some of these properties include the integral of an even function over a symmetric interval, using antiderivatives to evaluate definite integrals, and the behavior of an integral when the limits are the same.

For instance, an often-used property is that if a function is even, meaning that \(f(t) = f(-t)\), and the interval is symmetric about the origin, we can integrate from 0 to 'a' and then double the result, rather than integrating from '-a' to 'a'. Moreover, if the function is odd, meaning \(f(-t) = -f(t)\), then its definite integral over a symmetric interval is zero.

The solution to the exercise walks through how these properties can be applied to the different cases of \(f(t)\). In particular, knowing that the integral of a constant over an interval is that constant multiplied by the interval length simplifies case (c) tremendously. Applying the properties of definite integrals can simplify many IIT JEE problems significantly, which is why they are an essential part of the curriculum.