Question

даю 20 балов
помогите​

Answer 1

№ 4.1:

$1) \int\limits^2_ {-3}(2x-3) \, dx = \int\limits^2_ {-3}2x \, dx - \int\limits^2_ {-3}3 \, dx=2* \int\limits^2_ {-3}x \, dx - 3 \int\limits^2_ {-3} \, dx = \frac{2x^{2} }{2} - 3x = x^{2} -3x = x(x - 3) = 2(2-3) + 3 (-6 - 3) = -2 -27 = -29$

$3) \int\limits^0_{-2}(3x^{2}+10) \, dx = 3\int\limits^0_{-2}x^{2} \, dx + 10\int\limits^0_{-2} \, dx = \frac{3x^{3} }{3}+10x = x^{3} + 10x = x(x^{2} + 10)= 0 (0 +10) +2(4+10) = 0 + 28 = 28$

$5) \int\limits^4_2 {|x-3|} \, dx = \int\limits^3_2 {-(x-3)} \, dx + \int\limits^4_3 {x-3} \, dx= 3\int\limits^3_2 {} \, dx - \int\limits^3_2 {x} \, dx + \int\limits^4_3 {x} \, dx - 3\int\limits^4_3 {} \, dx = \frac{1}{2} + \int\limits^4_3 {x-3} \, dx = \frac{1}{2} + \frac{1}{2} =1$

№ 4.2:

$1) \int\limits^\frac{5\pi }{6} _\frac{\pi }{6} {cosx} \, dx = sinx = sin\frac{5\pi }{6} - sin\frac{\pi }{6} = \frac{1}{2} - \frac{1}{2} = 0$

$3) \int\limits^1_ {-1}(5x^{4} +6x^{2}) \, dx = 5\int\limits^1_ {-1}x^{4} \, dx + 6\int\limits^1_{-1}x^{2} \, dx = \frac{5x^{5} }{5} + \frac{6x^{3} }{3 } = x^{5} + 2x^{3} = 1^{5} + 2*1^{3} - (-1)^{5} + 2 * (-1)^{3} = 1 + 2 + 1 -2 = 2$