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homeworkhelpls
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i integrated y to get (1/3x^3 + 2x) with upper limit 5 / lower limit 2 but got 45 not 175 / 3
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Both Wolfram Alpha and I agree with you.homeworkhelpls said:View attachment 321610
i integrated y to get (1/3x^3 + 2x) with upper limit 5 / lower limit 2 but got 45 not 175 / 3
Just a notation tip. 1/3x^3 can be misread as 1/(3x^3) placing the x^3 in the denominator. To be precise, we can instead write (1/3)x^3 to ensure x^3 is in the numerator and not mistakenly placed in the denominator.homeworkhelpls said:View attachment 321610
i integrated y to get (1/3x^3 + 2x) with upper limit 5 / lower limit 2 but got 45 not 175 / 3
The answer is not 45 because when integrating to find the area under a curve, you are essentially finding the sum of infinitely small rectangles under the curve. Therefore, the answer will be a decimal or fraction, rather than a whole number like 45.
To integrate and find the area under a curve, you must first find the antiderivative of the function. Then, you can use the fundamental theorem of calculus to evaluate the integral at the given limits of integration.
The answer is in the form of 175/3 because it is the exact value of the integral. In some cases, the integral may not have a simplified decimal or whole number answer, so it is left in fraction form.
Integration is essentially finding the area under a curve, which can be thought of as the sum of infinitely small rectangles. It is used to solve problems involving rates of change and accumulation.
Integrating to find the area under a curve has many real-world applications, such as calculating the distance traveled by an object given its velocity function, finding the total revenue of a business over a certain time period, and determining the amount of medication in a patient's bloodstream over time.