Помогите решить. Студент 1 курс

Решите задачу:

$\huge \\ 1.\: S=\int_{2}^{5}x\mathrm dx={x^2\over2}|_{2}^{5}={1\over2}(25-4)={21\over2}=10,5\\ 2.\: \int_{-1}^{4}3x\mathrm dx=3\int_{-1}^{4}x\mathrm dx={3\over2}x^2|_{-1}^{4}={3\over2}(16-1)={3\over2}\cdot15={45\over2}=22,5\\\\ 3.\: \int_{-2}^{0}x^2\mathrm dx=\int_{0}^{2}x^2\mathrm dx\\ \int_{-2}^{0} x^2\mathrm dx={1\over3}x^3|_{-2}^{0}={1\over3}(0+8)={8\over3}\\ \int_{0}^{2} x^2\mathrm dx={1\over3}x^3|_{0}^{2}={1\over3}(8+0)={8\over3}$
$\\ 4.\: \int_{1}^{e}{1\over1+2x}\mathrm dx={1\over2}\int_{1}^{e}{1\over1+2x}\mathrm d(1+2x)={1\over2}\ln{|1+2x|}_{1}^{e}={1\over2}(\ln({1+2e})-\ln{3})={1\over2}\ln{1+2e\over3}$