MHB Finding Area under x^2-16 Curve: ∫_2^5

[markosheehan]

Find the area bound by the curve y=x^2-16 , the x-axis and the lines x=2 and x=5. i am trying to use the definite integral way ∫_2^5. but i am not getting the right answer. the right answer is 53/3

 

[I like Serena]

Find the area bound by the curve y=x^2-16 , the x-axis and the lines x=2 and x=5. i am trying to use the definite integral way ∫_2^5. but i am not getting the right answer. the right answer is 53/3

Those are 2 areas.
Taking the integral counts one negative and the other positive.
Both should be positive to find the area.

 

[MarkFL]

Yes, to find the area $A$ bounded by two functions ($f$ and $g$) over an interval $[a,b]$, we need to use:

$$A=\int_a^b \left|f(x)-g(x)\right|\,dx$$

Now, in the given problem we may define:

$$f(x)=x^2-16$$

$$g(x)=0$$

And so we then find:

$$\left|f(x)-g(x)\right|=\left|x^2-16\right|=\begin{cases}x^2-16, & |x|>4 \\[3pt] 16-x^2, & |x|\le4 \\ \end{cases}$$

With $(a,b)=(2,5)$, we may then write:

$$A=\int_2^4 16-x^2\,dx+\int_4^5 x^2-16\,dx$$