/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 4 Find the area between the curves... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the area between the curves on the given interval. $$y=e^{-x}, y=x^{2}, 1 \leq x \leq 4$$

Short Answer

Expert verified
The area between the curves \(y=e^{-x}\) and \(y=x^{2}\) on the interval \([1, 4]\) is approximately 25.336 square units.

Step by step solution

01

Identify the top and bottom functions

Determine which function lies above the other within the given interval. Plotting the functions or computing a few points can help determine this. In this case, \(y=e^{-x}\) is greater than \(y=x^{2}\) in the interval \([1, 4]\). Therefore, \(e^{-x}\) is the top function and \(x^{2}\) is the bottom function.
02

Set up the integral

Set up the integral to represent the area between the curves. The area between two curves from \(a\) to \(b\) is given by \[\int_a^b |f(x) - g(x)| dx\]. In this exercise, it's \[\int_{1}^{4} (e^{-x} - x^{2}) dx\].
03

Evaluate the integral

Evaluate the definite integral. The antiderivative of \(e^{-x}\) is \(-e^{-x}\) and the antiderivative of \(x^{2}\) is \(\frac{x^{3}}{3}\). Applying the fundamental theorem of calculus gives us \[ -e^{-x} - \frac{x^{3}}{3} \Biggr|_1^4 = (-e^{-4} - \frac{64}{3}) - (-e^{-1} - \frac{1}{3})\].
04

Calculate the numeric value

Finally, perform the subtraction to get the numeric area value. This would yield approximately 25.336.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Definite Integral
When studying the area between curves in calculus, the definite integral is a fundamental concept. It's used to calculate the exact area under a curve within a specific interval. Think of it as a tally of little rectangles that add up to the total area under a curve from one point to another.

In our exercise involving the curves defined by the functions y=e^{-x} and y=x^{2} between x=1 and x=4, setting up the definite integral \[\int_{1}^{4} (e^{-x} - x^{2}) dx\] involves subtracting the function representing the lower curve (x^{2}) from the function representing the upper curve (e^{-x}), across the specified interval. This operation allows us to find the total 'accumulated' area. The magic of the definite integral is that regardless of the curve's shape, it gives us a precise measurement, a numeric answer representing the total area between these two functions.
Exponential Functions
Exponential functions, like the one seen in the exercise with y=e^{-x}, are critical in calculus and across various scientific fields. These functions grow (or decay) at rates proportional to their current value, creating a distinctive, continuous, and smooth curve.

An important characteristic of the exponential function e^{-x} is that it never touches the x-axis but gets infinitely close, heading towards zero as x increases. This behavior is crucial when calculating the area under the curve using integration, as seen in our exercise, where e^{-x} contributes to the boundary around the area we're measuring. Understanding the nature of exponential functions helps to grasp the logic behind the subtraction in the integral setup, ensuring we account for the correct region in the space between the curves.
Fundamental Theorem of Calculus
The final piece of the puzzle is the Fundamental Theorem of Calculus, a towering concept that bridges the gap between derivative and integral. It essentially tells us that integration and differentiation are inverse processes.

When applying the fundamental theorem to our exercise, it governs how we move from the set-up of the integral to finding the actual area value. After setting up the integral \[\int_{1}^{4} (e^{-x} - x^{2}) dx\], the theorem guides us to find the antiderivatives, calculate the expressions at the upper and lower bound, and subtract these values to reveal the area between the curves. In essence, it legitimizes the operation by guaranteeing that if we correctly find the cumulative total represented by the integral, we're indeed calculating the exact area we're interested in, here approximated as 25.336.

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