Fluency exercise

## Solution

Can you find the missing positive numbers $a$ to $d$?

We have four integrals to look at, but only two of them contain just one of the letters—(1) and (3).

$\text{(1)} \ \int_a^5 10x+3 \, dx = 114.$

We can see that the value the integral takes is positive, so we have an area formed between $10x+3$ and the $x$-axis.

This can only happen when $0<a<5$, since $a$ is also positive. We now have a good feeling for the sort of value for $a$ we are looking for.

If we perform the integration we get the following,

$\text{(1)} \ \int_a^5 10x+3 \, dx = \left[5x^2+3x \right]^5_a = (5\times 5^2+3\times 5)-(5a^2+3a) = 114.$

Simplifying the above we find that,

\begin{align*} 140-5a^2-3a &=114\\ \implies 5a^2+3a-26 &=0 \end{align*}

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.

Which integral could we tackle next?