giúp vơiiiiiiiiiiiiii an Luyện tập 3+4. Tính các TP sau : TH2: "),$a/~\int^{2\pi}_0(2x+\cos x)dx.$ $h/~\int^1_0(1+\sin x)dx$,$b)~\int^2_1(3^x-\frac32)dx$ $i)~\int^2_{-1}|2x+1|dx$,$c/~\int^1_\pi(\frac1{\cos^2x}-\frac1{\sin^2x})dx$,$d/~\int^3_{10}|2x-3|dx$,$e/~\int^3_0(3x-1)^2dx.$,$g/~\int^1_0(e^{2x}+3x^2)dx$

Question

Accepted Answer

a) Ta có:
$$ 
\int^{2\pi}_0(2x + \cos x) \, dx = \left[ x^2 + \sin x \right]^{2\pi}_0 = (4\pi^2 + \sin 2\pi) - (0 + \sin 0) = 4\pi^2.
$$

b) Ta có:
$$ 
\int^2_1 \left( 3^x - \frac{3}{2} \right) \, dx = \left[ \frac{3^x}{\ln 3} - \frac{3}{2}x \right]^2_1 = \left( \frac{3^2}{\ln 3} - \frac{3}{2} \cdot 2 \right) - \left( \frac{3^1}{\ln 3} - \frac{3}{2} \cdot 1 \right) = \frac{9}{\ln 3} - 3 - \frac{3}{\ln 3} + \frac{3}{2} = \frac{6}{\ln 3} - \frac{3}{2}.
$$

c) Ta có:
$$ 
\int^{\pi/3}_{\pi/4} \left( \frac{1}{\cos^2 x} - \frac{1}{\sin^2 x} \right) \, dx = \int^{\pi/3}_{\pi/4} (\tan^2 x - \cot^2 x) \, dx = \left[ \tan x + \cot x \right]^{\pi/3}_{\pi/4} = (\sqrt{3} + \frac{1}{\sqrt{3}}) - (1 + 1) = \sqrt{3} + \frac{1}{\sqrt{3}} - 2.
$$

d) Ta có:
$$ 
\int^3_0 |2x - 3| \, dx = \int^3_0 (3 - 2x) \, dx + \int^3_{3/2} (2x - 3) \, dx = \left[ 3x - x^2 \right]^3_0 + \left[ x^2 - 3x \right]^3_{3/2} = (9 - 9) + (9 - 9) - \left( \frac{9}{4} - \frac{9}{2} \right) = \frac{9}{4}.
$$

e) Ta có:
$$ 
\int^3_0 (3x - 1)^2 \, dx = \int^3_0 (9x^2 - 6x + 1) \, dx = \left[ 3x^3 - 3x^2 + x \right]^3_0 = (81 - 27 + 3) - (0) = 57.
$$

f) Ta có:
$$ 
\int^{\pi/2}_0 (1 + \sin x) \, dx = \left[ x - \cos x \right]^{\pi/2}_0 = \left( \frac{\pi}{2} - 0 \right) - (0 - 1) = \frac{\pi}{2} + 1.
$$

g) Ta có:
$$ 
\int^1_0 (e^{2x} + 3x^2) \, dx = \left[ \frac{e^{2x}}{2} + x^3 \right]^1_0 = \left( \frac{e^2}{2} + 1 \right) - \left( \frac{1}{2} + 0 \right) = \frac{e^2}{2} + 1 - \frac{1}{2} = \frac{e^2}{2} + \frac{1}{2}.
$$

h) Ta có:
$$ 
\int^2_{-1} |2x + 1| \, dx = \int^{-1/2}_{-1} (-2x - 1) \, dx + \int^2_{-1/2} (2x + 1) \, dx = \left[ -x^2 - x \right]^{-1/2}_{-1} + \left[ x^2 + x \right]^2_{-1/2} = \left( -\frac{1}{4} + \frac{1}{2} \right) - (1 - 1) + (4 + 2) - \left( \frac{1}{4} - \frac{1}{2} \right) = \frac{1}{4} + 6 - \frac{1}{4} = 6.
$$

Đáp số:
a) $ 4\pi^2 $
b) $ \frac{6}{\ln 3} - \frac{3}{2} $
c) $ \sqrt{3} + \frac{1}{\sqrt{3}} - 2 $
d) $ \frac{9}{4} $
e) $ 57 $
f) $ \frac{\pi}{2} + 1 $
g) $ \frac{e^2}{2} + \frac{1}{2} $
h) $ 6 $