Question

Question #ad67f

Answer 1

The region has an area of `500/3` square units.

Explanation:

I'm going to start at the start of this problem and work my way to the end methodically so that you know exactly what is going on.

We're going to need to find the equation of the tangent line, so we differentiate the parabola's equation (call it `f(x)`) to `f'(x) =8x` by the power rule.

The slope of the tangent is therefore `f'(5) = 8(5) = 40`.

The equation of the tangent is therefore:

`y- 100 = 40(x - 5)`

`y- 100 = 40x - 200`

`y = 40x - 100`

We now compare the two graphs, letting the tangent line be `g(x)`. Whenever `x < 5`, `f(x) > g(x)`. The enclosed area is in `0 ≤ x ≤ 5`, so `f(x)`, the parabola, is the top graph.

We now set up our integral:

`=>int_0^5 4x^2 - (40x - 100) dx`

`=>int_0^5 4x^2 - 40x + 100 dx`

`=>int_0^5 4(x^2 - 10x + 25)dx`

`=>4int_0^5 x^2 - 10x + 25dx`

Now integrate using `intx^ndx = x^(n + 1)/(n+ 1)`. Since this is a definite integral, we can forget about the constant of integration, `C`, that would generally be added onto the computed indefinite integral.

`=>4[1/3x^3 - 5x^2 + 25x]_0^5`

This can be evaluated using the second fundamental theorem of calculus, which states that `int_a^b F(x) = f(b) - f(a)`, where `f'(x) = F(x)`.

`=>4[1/3(5)^3 - 5(5)^2 + 25(5) - (1/3(0)^3 + 5(0)^2 + 25(0))]`

`=>4[125/3 - 125 + 125]`

`=>500/3`

Hopefully this helps!