The figure shows the graph of 𝑓 of
𝑥 equals a quarter 𝑥 minus two squared 𝑥 plus one. Calculate the area of the shaded
region, giving your answer as a fraction.
To calculate the area between a
curve and the 𝑥-axis, we need to integrate that function with respect to 𝑥. In this case, we’ll be integrating
the function a quarter 𝑥 minus two squared 𝑥 plus one with respect to 𝑥. And that’s between the limits two
and negative one, since those are the 𝑥-values which represent the upper and lower
limits of the area required.
We do, however, need to be a little
bit careful since integrating a function whose area lies underneath the 𝑥-axis will
give a negative value. In this case so, we’re purely
interested in the area above the 𝑥-axis. So, we’re fine just to integrate
here. And the easiest way to integrate
this function, in my opinion, is to distribute the parentheses. We could use a substitution of 𝑢
equals 𝑥 minus two, but it’s simple enough to just read these parentheses. So, let’s go ahead and do that.
We begin by distributing 𝑥 minus
two all squared. By remembering that we need to
write this as the product of two brackets, 𝑥 minus two and 𝑥 minus two. 𝑥 multiplied by 𝑥 is 𝑥
squared. We multiply the outer terms. We get negative two 𝑥. Then we multiply the inner terms,
and we get negative two 𝑥 again. Then multiplying the last terms, we
get positive four. So, we can see 𝑥 minus two all
squared is equal to 𝑥 squared minus four 𝑥 plus four.
We’re then going to multiply this
by 𝑥 plus one, being really careful to multiply each term in the first parenthesis
by each term in the second. One way to do this is to use the
grid method. 𝑥 multiplied by 𝑥 squared is 𝑥
cubed. And 𝑥 squared multiplied by one is
𝑥 squared. 𝑥 multiplied by negative four 𝑥
is negative four 𝑥 squared. And negative four 𝑥 multiplied by
one is negative four 𝑥. 𝑥 multiplied by four is four
𝑥. And four multiplied by one is
four. If we collect like terms, we can
see that 𝑥 minus two all squared multiplied by 𝑥 plus one is 𝑥 cubed minus three
𝑥 squared plus four.
So, we need to actually integrate a
quarter 𝑥 cubed minus three 𝑥 squared plus four with respect to 𝑥 between those
limits two and negative one we said earlier. In fact, whenever we have a
function in 𝑥 is being multiplied by some constant, it can be simpler to bring this
constant outside of the integration sign. This is allowed. We will simply end up multiplying
by a quarter at the end, rather than at the beginning of our calculations.
Let’s integrate 𝑥 cubed minus
three 𝑥 squared plus four. To integrate 𝑥 cubed, we add one
to the exponent. That gives us 𝑥 to the power of
four. We then divide through by the value
of that new exponent. So, 𝑥 cubed becomes 𝑥 to the
power of four divided by four. Negative three 𝑥 squared becomes
negative three 𝑥 cubed divided by three, which is simply negative 𝑥 cubed. And four integrates to become four
𝑥. Remember we do not need the
constant of integration when integrating between two limits.
We’ll just need to evaluate between
the limits two and negative one. So, we substitute two and negative
one into our expression and find the difference between them. Substituting two in, we get two to
the power of four divided by four minus two cubed plus four multiplied by two, which
is four. And the second part is negative one
to the power of four divided by four minus negative one cubed plus four multiplied
by negative one. And that simplifies to negative 11
over four. And if we type all of this into our
calculator, we get an answer of 27 over 16. And so, we can see that the shaded
area is 27 over 16 square units.