Chapter 4 - Integrals - Review - Exercises - Page 349: 5

$3$

Since $4$ is between $0$ and $6$ then by the Integral of the Function over Adjacent Intervals it follows: $\int_{0}^{6} f(x)dx=\int_{0}^{4} f(x)dx+\int_{4}^{6} f(x)dx$ Substitute each definite integral by its value: $10=7+\int_{4}^{6} f(x)dx$ $10-7=\int_{4}^{6} f(x)dx$ $3=\int_{4}^{6} f(x)dx$ So the answer is: $\int_{4}^{6} f(x)dx=3$