If $\int_3^5 p(x)\,dx=3$ and $\int_8^5 p(x)\,dx=1$, then what is the value of $\int_3^8 (p(x)+1)\,dx$?

Key concept: This problem combines two basic ideas: reversing the limits of integration and adding integrals over adjacent intervals.

Step by step

Start by rewriting the integral with limits in increasing order.

Given:

$\int_3^5 p(x)\,dx=3$

and

$\int_8^5 p(x)\,dx=1.$

Reverse the second integral:

$\int_8^5 p(x)\,dx=-\int_5^8 p(x)\,dx=1,$

so

$\int_5^8 p(x)\,dx=-1.$

Now add the two adjacent intervals:

$\int_3^8 p(x)\,dx=\int_3^5 p(x)\,dx+\int_5^8 p(x)\,dx=3+(-1)=2.$

Next, integrate the constant $1$:

$\int_3^8 1\,dx=8-3=5.$

Therefore,

$\int_3^8 (p(x)+1)\,dx=\int_3^8 p(x)\,dx+\int_3^8 1\,dx=2+5=7.$

Answer: $7$

Pitfall alert

Be careful with reversed limits: $\int_8^5 p(x)\,dx$ is the negative of $\int_5^8 p(x)\,dx$. Another common mistake is forgetting to split $\int_3^8 (p(x)+1)\,dx$ into two separate integrals.

Try different conditions

If the constant were $c$ instead of $1$, then $\int_3^8 (p(x)+c)\,dx = \int_3^8 p(x)\,dx + 5c$. If the second given integral had limits $5$ to $8$ instead of $8$ to $5$, the sign would change and the total would be different.

Further reading

reversing limits, additivity of integrals, constant integrand