Find the area using integration.

[turbokaz]

Homework Statement

Find the area bounded by the graph f(x)=2x^2, point P(1, f(1)), and the x axis.

Homework Equations

The Attempt at a Solution

Point P is a point on a line that is tangent to y=2x^2. So I used this to figure that the line is 4x-2. So now there are two functions, y=2x^2 and y=4x-2, with the two functions intersecting at (1,2). I did the integral from 0 to 1 of top minus bottom, which is 2x^2-4x+2. Taking the antiderivatives, I get 2x^3/3 -2x^2 +2x. From 0 to 1, I get an answer of 2/3. But that isn't correct apparently. The answer choices are 1/8, 1/6, 1/2, 1/3 and 1/4. Where am I wrong?

 

[Mark44]

Homework Statement

Find the area bounded by the graph f(x)=2x^2, point P(1, f(1)), and the x axis.

Your problem description is not as good as it should be. One of the boundaries of the region is the tangent line through (1, f(1)).

Homework Equations

The Attempt at a Solution

Point P is a point on a line that is tangent to y=2x^2. So I used this to figure that the line is 4x-2. So now there are two functions, y=2x^2 and y=4x-2, with the two functions intersecting at (1,2). I did the integral from 0 to 1 of top minus bottom, which is 2x^2-4x+2. Taking the antiderivatives, I get 2x^3/3 -2x^2 +2x. From 0 to 1, I get an answer of 2/3. But that isn't correct apparently. The answer choices are 1/8, 1/6, 1/2, 1/3 and 1/4. Where am I wrong?

You aren't taking into full account the tangent line. Over part of the interval, the x-axis is the bottom boundary of the region. Over the other part of the interval, the tangent line is the bottom boundary. You need two integrals, since the integrand is different over the two intervals.

A piece of information that you have neglected is the x-intercept of your tangent line.

BTW, I get one of the values you list as potential answers.