Applications Of Integrals

We are given the curve $y%20%3D%20x(2%20-%20x)$ and the line $y%20%3D%20x$ . To find the area bounded by the curve and the line, we first need to find the points of intersection of these two functions.
Step 1: Find the points of intersection Set $x(2%20-%20x)%20%3D%20x$ to find the points of intersection: $x(2%20-%20x)%20%3D%20x$ $2x%20-%20x%5E2%20%3D%20x$ $2x%20-%20x%5E2%20-%20x%20%3D%200$ $-x%5E2%20%2B%20x%20%3D%200$ $x(x%20-%201)%20%3D%200$ Thus, $x%20%3D%200$ and $x%20%3D%201$ are the points of intersection.
Step 2: Set up the integral for the area The area between the curve and the line from $x%20%3D%200$ to $x%20%3D%201$ is given by the integral: $%5Ctext%7BArea%7D%20%3D%20%5Cint_0%5E1%20%5Cleft%5B%20x(2%20-%20x)%20-%20x%20%5Cright%5D%20dx$ Simplify the integrand: $x(2%20-%20x)%20-%20x%20%3D%202x%20-%20x%5E2%20-%20x%20%3D%20x%20-%20x%5E2$ Thus, the area is: $%5Ctext%7BArea%7D%20%3D%20%5Cint_0%5E1%20(x%20-%20x%5E2)%20dx$
Step 3: Compute the integral Integrate $x%20-%20x%5E2$ : $%5Cint%20(x%20-%20x%5E2)%20dx%20%3D%20%5Cfrac%7Bx%5E2%7D%7B2%7D%20-%20%5Cfrac%7Bx%5E3%7D%7B3%7D$ Now evaluate the definite integral: $%5Ctext%7BArea%7D%20%3D%20%5Cleft%5B%20%5Cfrac%7Bx%5E2%7D%7B2%7D%20-%20%5Cfrac%7Bx%5E3%7D%7B3%7D%20%5Cright%5D_0%5E1%20%3D%20%5Cleft(%20%5Cfrac%7B1%5E2%7D%7B2%7D%20-%20%5Cfrac%7B1%5E3%7D%7B3%7D%20%5Cright)%20-%20%5Cleft(%20%5Cfrac%7B0%5E2%7D%7B2%7D%20-%20%5Cfrac%7B0%5E3%7D%7B3%7D%20%5Cright)$ $%5Ctext%7BArea%7D%20%3D%20%5Cleft(%20%5Cfrac%7B1%7D%7B2%7D%20-%20%5Cfrac%7B1%7D%7B3%7D%20%5Cright)$ $%5Ctext%7BArea%7D%20%3D%20%5Cfrac%7B3%7D%7B6%7D%20-%20%5Cfrac%7B2%7D%7B6%7D%20%3D%20%5Cfrac%7B1%7D%7B6%7D$

The correct option is (A) : $%5Cfrac%7B1%7D%7B6%7D$

We are given the differential equation:

%5Cfrac%7Be%5Ey%7D%7Bdx%7D%20%3D%20x%20%2B%202

First, rewrite it in the form:

e%5Ey%20%5Cfrac%7Bdy%7D%7Bdx%7D%20%3D%20x%20%2B%202

Now, separate the variables

y

and

x

e%5Ey%20dy%20%3D%20(x%20%2B%202)%20dx

Integrate both sides:

%5Cint%20e%5Ey%20%5C%2C%20dy%20%3D%20%5Cint%20(x%20%2B%202)%20%5C%2C%20dx

The integral of

e%5Ey

with respect to

y

e%5Ey

, and the integral of

(x%20%2B%202)

with respect to

x

is:

%5Cint%20(x%20%2B%202)%20%5C%2C%20dx%20%3D%20%5Cfrac%7B(x%20%2B%202)%5E2%7D%7B2%7D%20%2B%20C

Thus, the solution is:

e%5Ey%20%3D%20%5Cfrac%7B(x%20%2B%202)%5E2%7D%7B2%7D%20%2B%20C

Solving for

y

, we get:

y%20%3D%20%5Clog(x%20%2B%202)%20%2B%20C

Thus, the correct answer is option (B):

y%20%3D%20(x%20%2B%202)%20%5Clog(x%20%2B%202)%20-%20x%20%2B%20C

Thus, the correct answer is option (B).

About Applications Of Integrals - KEAM

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Applications Of Integrals

High-Yield Trend

Chapter Questions 3 MCQs

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About Applications Of Integrals - KEAM

Frequently Asked Questions

Why focus on Applications Of Integrals PYQs?

How to best use this analysis?

Chapter Questions
3 MCQs