1. The problem asks to evaluate the definite integrals \(\int_{-1}^2 f(x)\,dx\), \(\int_2^4 f(x)\,dx\), and \(\int_{-1}^7 f(x)\,dx\) by interpreting them as areas under the curve of \(f(x)\). 2. Important rule: The definite integral \(\int_a^b f(x)\,dx\) represents the net area between the graph of \(f(x)\) and the x-axis from \(x=a\) to \(x=b\). Areas above the x-axis count as positive, and areas below count as negative. 3. Given \(\int_{-1}^7 f(x)\,dx = 1\), we use this to find the other integrals by breaking the interval \([-1,7]\) into \([-1,2]\), \([2,4]\), and \([4,7]\). 4. From the graph description: - From \(x=-1\) to \(x=2\), the graph goes from about 1 down to -2, so the area includes positive and negative parts. - From \(x=2\) to \(x=4\), the graph forms a semicircular arc above the x-axis. - From \(x=4\) to \(x=7\), the graph increases linearly and then stays constant. 5. Let \(A_1 = \int_{-1}^2 f(x)\,dx\), \(A_2 = \int_2^4 f(x)\,dx\), and \(A_3 = \int_4^7 f(x)\,dx\). 6. Since \(A_1 + A_2 + A_3 = 1\), we need to find \(A_1\) and \(A_2\) to solve for \(A_3\). 7. Approximate \(A_2\): The semicircle from \(x=2\) to \(x=4\) has radius 1 (since diameter is 2 units). Area of semicircle = \(\frac{1}{2} \pi r^2 = \frac{1}{2} \pi (1)^2 = \frac{\pi}{2} \approx 1.5708\). 8. Approximate \(A_1\): The area from \(x=-1\) to \(x=2\) includes a triangle above x-axis and a triangle below. The positive area is roughly a triangle with base 1.5 and height 1, area \(= \frac{1}{2} \times 1.5 \times 1 = 0.75\). The negative area is roughly a triangle with base 1.5 and height 2, area \(= \frac{1}{2} \times 1.5 \times 2 = 1.5\). Net area \(A_1 = 0.75 - 1.5 = -0.75\). 9. Now solve for \(A_3\): $$A_3 = 1 - A_1 - A_2 = 1 - (-0.75) - 1.5708 = 1 + 0.75 - 1.5708 = 0.1792$$ 10. Final answers: - \(\int_{-1}^2 f(x)\,dx = -0.75\) - \(\int_2^4 f(x)\,dx = \frac{\pi}{2} \approx 1.5708\) - \(\int_{-1}^7 f(x)\,dx = 1\) (given) These values are consistent with the graph and the given total integral.