And once again, even if you didn't see the graph, you would know that because in both cases you're getting the definite integral of f of x, dx, and our upper bound here is our lower bound here. So once again, we're able to merge the integrals. And what is this going to be equal to? Well, we have this area here, which is five, and then we have this area, which is six. That was given to us. But since it's below the x-axis and above our curve, when we evaluate it as a definite integral, it would evaluate as a negative six. So this is going to be five plus negative six, which is equal to negative one, and we're done.