The prime factor decomposition of \(26\) is \(2\times 13\). Trying the possible combinations leads us to \((5x+13)(x-2)\). Given that \(a\) is positive, we must have \(a=2\).

Is this sensible?

Well, we thought above that \(0<a<5\) and we have found a value that fits with this.

This equation contains all four of the unknowns so is best kept til last. As before, we start by rewriting the integrand and substituting in the values for \(a\), \(b\) and \(c\).

We can also substitute our values for \(a\), \(b\) and \(c\) into the right hand side of the original equation to give \(ab^ad^a=2\times 3^2\times d^2\).

\[\int^{12}_6 x^2-x \, dx=18d^2.\]

Performing the integration and solving a quadratic equation in \(d\), we find that \(d=5\).

The area formed between the \(x\)-axis and the lines \(x=b\) and \(x=d\), and the curve \(y=(x-2a)(x+1)\) is \(\dfrac{cd}{a(a+b)}\).

To check that we have found the correct values for \(a\), \(b\), \(c\) and \(d\) we can evaluate the integral between the given \(x\) values,

and check it is equal to \(\dfrac{cd}{a(a+b)}=5\).

But we are not told about the integral, we are told that the area formed between the curve and the \(x\)-axis is \(5\). This means we need to be careful if our curve crosses the \(x\)-axis.

Sketching the curve \(y=(x-4)(x+1)=x^2-3x-4\) reveals that the curve crosses the \(x\)-axis at \(x=4\). So we must separately calculate the areas between the \(x\) values \(3\) and \(4\), and \(4\) and \(5\).

We can thus split our integral up into two

\[\int_3^4 x^2-3x-4 \, dx,\]

which will give the negative value \(-\dfrac{13}{6}\) and

\[\int_4^5 x^2-3x-4 \, dx,\]

which will give the positive value \(\dfrac{17}{6}\).

Therefore, the total area formed between the curve, the lines \(x=3\) and \(x=5\), and the \(x\)-axis is \(\dfrac{13}{6} + \dfrac{17}{6} = 5\).