(1) y=8x2+4y = \dfrac{8}{x^2+4}y=x2+48 x=2yx = 2yx=2y x=0x = 0x=0 ∫028x2+4 dx−∫0212x dx=[8⋅12tan−1x2]02−[14x2]02=4⋅π4−1=π−1\int_0^2 \frac{8}{x^2+4}\,dx - \int_0^2 \frac{1}{2}x\,dx = \left[8\cdot\frac{1}{2}\tan^{-1}\frac{x}{2}\right]_0^2 - \left[\frac{1}{4}x^2\right]_0^2 = 4\cdot\frac{\pi}{4}-1=\pi-1∫02x2+48dx−∫0221xdx=[8⋅21tan−12x]02−[41x2]02=4⋅4π−1=π−1 (2) y=8x2+4y = \dfrac{8}{x^2+4}y=x2+48 y=x24y = \dfrac{x^2}{4}y=4x2 ∫−228x2+4 dx−∫−22x24 dx=2{∫028x2+4−x24 dx}=2[4tan−1x2−x312]02=2(4⋅π4−23)=2π−43\int_{-2}^2 \frac{8}{x^2+4}\,dx - \int_{-2}^2 \frac{x^2}{4}\,dx = 2\left\{\int_0^2 \frac{8}{x^2+4}-\frac{x^2}{4}\,dx\right\} = 2\left[4\tan^{-1}\frac{x}{2}-\frac{x^3}{12}\right]_0^2 = 2\left(4\cdot\frac{\pi}{4}-\frac{2}{3}\right)=2\pi-\frac{4}{3}∫−22x2+48dx−∫−224x2dx=2{∫02x2+48−4x2dx}=2[4tan−12x−12x3]02=2(4⋅4π−32)=2π−34