Nguyên hàm - Bài kiểm tra
Lời giải:
\( F(x) = \int \dfrac{3}{\sqrt{x+2}}\,dx = 3 \int (x+2)^{-\frac{1}{2}}\,dx = 3 \cdot \dfrac{(x+2)^{\frac{1}{2}}}{\frac{1}{2}} + c \)
\( = 6\sqrt{x+2} + c \)
Chọn: C
Page 1
Chọn: B
Page 2
Lời giải:
Ta có: \( f(x) = \dfrac{x - 2}{x + 1} = \dfrac{x + 1 - 3}{x + 1} = 1 - \dfrac{3}{x + 1} \)
\( Rightarrow F(x) = \int f(x)\,dx = x - 3\ln|x+1| + c \)
Chọn: A
Page 3
Lời giải:
\( F(x) = \int 2\cos(2x)\,dx = \sin(2x) + C \)
\( F(\pi) = \sin(2\pi) + C = 0 + C = 1 \Rightarrow C = 1 \)
Vậy \( F(x) = \sin(2x) + 1 \Rightarrow F(0) = \sin(0) + 1 = 1 \)
Chọn: C
Page 4
Lời giải:
\( f'(x) = \dfrac{1}{x^2 \sqrt{x}} = \dfrac{1}{x^2 \cdot x^{\frac{1}{2}}} = \dfrac{1}{x^{2 + \frac{1}{2}}} = \dfrac{1}{x^{\frac{5}{2}}} = x^{-\frac{5}{2}} \)
\( f(x) = \int f'(x)\,dx = \int x^{-\frac{5}{2}}\,dx = \dfrac{x^{-\frac{5}{2} + 1}}{-\frac{5}{2} + 1} + C = \dfrac{x^{-\frac{3}{2}}}{-\frac{3}{2}} + C = -\dfrac{2}{3}x^{-\frac{3}{2}} + C \)
\( f(x) = -\dfrac{2}{3x\sqrt{x}} + C \)
\( f(1) = 2 \Rightarrow -\dfrac{2}{3} + C = 2 \Rightarrow C = \dfrac{8}{3} \)
\( f(x) = -\dfrac{2}{3\sqrt{x}} + \dfrac{8}{3} \Rightarrow f(4) = -\dfrac{1}{12} + \dfrac{8}{3} = \dfrac{31}{12} \)
Chọn: B
Page 5
Lời giải:
\( f'(x) = \dfrac{2x + 3}{x - 2} = \dfrac{2(x - 2) + 7}{x - 2} = 2 + \dfrac{7}{x - 2} \)
\( f(x) = \int f'(x)\,dx = \int \left(2 + \dfrac{7}{x - 2} \right) dx = 2x + 7\ln(x - 2) + C \)
\( f(3) = 6 + C = 7 \Rightarrow C = 1 \)
\( f(x) = 2x +1 + 7\ln(x - 2) \)
\( \Rightarrow f(5) = 11 + 7\ln 3 \)
Chọn: D
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Tính \( f(3) \)
Lời giải:
\( f'(x) \) là một hàm số bậc hai có bảng biến thiên như hình vẽ:
\( \Rightarrow f'(x) = (x-1)(x-3)\)
\( \Rightarrow f'(x) = x^2 - 4x + 3 \)
\( f(x) = \int f'(x)\,dx = \dfrac{x^3}{3} - 2x^2 + 3x + C \)
\( f(0) = 1 \Rightarrow C = 1\)
\( \Rightarrow f(x) = \dfrac{x^3}{3} - 2x^2 + 3x + 1 \)
\( f(3) = 9 - 18 + 9 + 1 = 1 \)
Chọn: D
Page 7
Lời giải:
Vì \( (\cos x)' = -\sin x\)
\(\Rightarrow F(x) = \int \dfrac{\sin x}{1 + 3\cos x} dx = -\dfrac{1}{3} \int \dfrac{-3\sin x}{1 + 3\cos x} dx = -\dfrac{1}{3} \ln|1 + 3\cos x| + C\)
\( F\left( \dfrac{\pi}{2} \right) = 2 \Leftrightarrow c = 2 \)
Vậy \( F(x) = -\dfrac{1}{3} \ln|1 + 3\cos x| + 2 \)
\( \Rightarrow F(0) = -\dfrac{1}{3} \ln(4 + 2 ) = -\dfrac{2}{3} \ln 2 + 2 \)
Chọn: B
Page 8
Lời giải:
\( f'(x) = \dfrac{1}{x^2 - 4x + 3} = \dfrac{1}{(x - 1)(x - 3)} = \dfrac{1}{2}( \dfrac{1}{x - 3} )- \dfrac{1}{2}( \dfrac{1}{x - 1})\)
\( f(x) = \int f'(x)\,dx = \int \left(\dfrac{1}{2}( \dfrac{1}{x - 3} )- \dfrac{1}{2}( \dfrac{1}{x - 1}) \right) dx \)
\( f(x) = \dfrac{1}{2} \ln|x - 3| - \dfrac{1}{2} \ln|x - 1| + C \)
\(= \dfrac{1}{2} \ln \left| \dfrac{x - 3}{x - 1} \right| + C \)
Biết \( f(2) = 0 \Rightarrow C = 0 \)
Vậy \( f(x) = \dfrac{1}{2} \ln \left| \dfrac{x - 3}{x - 1} \right| \)
\( f(4) = \dfrac{1}{2} \ln \left( \dfrac{1}{3} \right) = -\dfrac{1}{2} \ln 3 \)
Chọn: A
Page 9
Lời giải:
\( f(x) = \int f'(x) dx= \int \dfrac{1}{1 + \cos x} dx = \int \dfrac{1}{2\cos^2\left( \dfrac{x}{2} \right)} dx = \tan\left( \dfrac{x}{2} \right) + C \)
\( f\left( \dfrac{\pi}{2} \right) = \tan\left( \dfrac{\pi}{4} \right) + C = 1 \Rightarrow C = 0 \)
Vậy \(f(x) = tan (\dfrac{x}{2})\)
\( \Rightarrow f \left( \dfrac{\pi}{3} \right) = \tan\left( \dfrac{\pi}{6} \right) = \dfrac{\sqrt{3}}{3} \)
Chọn: B
Page 10
Lời giải:
Đặt \( u = 2x \Rightarrow du = 2dx \Rightarrow dx = \dfrac{1}{2} du \)
Khi \( x = 0 \Rightarrow u = 0 \), \( x = 1 \Rightarrow u = 2 \)
\( P = \int_0^1 f(2x)\,dx = \dfrac{1}{2} \int_0^2 f(u)\,du = \dfrac{1}{2} \left( \int_0^{10} f(u)\,du - \int_2^{10} f(u)\,du \right) = \dfrac{1}{2}(7 - 1) = 3 \)
Chọn: D
Page 11
Lời giải:
Đặt \( u = 2x + 1 \Rightarrow du = 2dx \Rightarrow dx = \dfrac{1}{2} du \)
\(
\begin{cases}
x = 1 \Rightarrow u = 3 \\
x = 3 \Rightarrow u = 7
\end{cases}
\)
\( \int_1^3 f(2x + 1)\,dx = 4 \Leftrightarrow \int_3^7 f(u)\dfrac{1}{2}\,du = 4 \)
\(\Rightarrow \int_3^7 f(u)\,du = 8 \)
Chọn: B
Page 12
Lời giải:
\(
\int_{1}^{5} f(2x - 5)\,dx
\)
Đặt \( u = 2x - 5 \Rightarrow du = 2\,dx \)
\(
\begin{cases}
x = 1 \Rightarrow u = -3 \\
x = 5 \Rightarrow u = 5
\end{cases}
\)
\( I = \int_1^5 f(2x - 5)\,dx = \dfrac{1}{2} \int_{-3}^5 f(u)\,du = \dfrac{1}{2} \cdot 10 = 5 \)
Chọn: B
Page 13
Lời giải:
Đặt \(
\begin{cases}
u = \ln x \\
dv = (4x + 2)\,dx
\end{cases}
\Rightarrow
\begin{cases}
du = \dfrac{1}{x}\,dx \\
v = 2x^2 + 2x
\end{cases}
\)
\(
\int_{2}^{3} (4x + 2) \ln x\, dx = \left( 2x^2 + 2x \right) \ln x \big|_{2}^{3} - \int_{2}^{3} (2x + 2)\, dx
\)
\(
= 24 \ln 3 - 12 \ln 2 - (x^2 + 2x)\big|_{2}^{3}
\)
\(
= 24 \ln 3 - 12 \ln 2 - 7
\)
\( \Rightarrow
\begin{cases}
a = 24 \\
b = -12 \\
c = -7
\end{cases}
\Rightarrow a + b + c = 5
\)
Chọn: B
Page 14
Lời giải:
Xét \(
\int_{0}^{3} f(\sqrt{x + 1})\, dx
\)
Đặt: \( u = \sqrt{x + 1} \Rightarrow du = \dfrac{1}{2\sqrt{x + 1}}\, dx \Rightarrow dx = 2u\, du \)
\(
\begin{cases}
x = 0 \Rightarrow u = 1 \\
x = 3 \Rightarrow u = 2
\end{cases}
\)
\(
8 = \int_{0}^{3} f(\sqrt{x + 1})\, dx = \int_{1}^{2} 2u f(u)\, du
\)
\(
\Rightarrow \int_{1}^{2} u f(u)\, du = 4
\)
Chọn: D
Page 15
Lời giải:
\( I = \int_{0}^{2} \dfrac{5x + 7}{x^2 + 3x + 2}\, dx = \int_{0}^{2} \left( \dfrac{5}{2} \cdot \dfrac{2x + 3 + 7 - \dfrac{15}{2}}{x^2 + 3x + 2} \right)\, dx \)
\( = \dfrac{5}{2} \int_{0}^{2} \dfrac{2x + 3}{x^2 + 3x + 2}\, dx
- \dfrac{1}{2} \int_{0}^{2} \dfrac{1}{(x + 1)(x + 2)}\, dx \)
\( = \dfrac{5}{2} \ln(x^2 + 3x + 2)\big|_{0}^{2}
- \dfrac{1}{2} \int_{0}^{2} \left( \dfrac{1}{x + 1} - \dfrac{1}{x + 2} \right)\, dx \)
\( = \dfrac{5}{2} (\ln12 - \ln2)
- \dfrac{1}{2} \left( \ln\left| \dfrac{x + 1}{x + 2} \right|\big|_{0}^{2} \right) \)
\( = \dfrac{5}{2} \ln(6)
- \dfrac{1}{2} \left( \ln\left( \dfrac{3}{4} \right) - \ln\left( \dfrac{1}{2} \right) \right) \)
\( = \dfrac{5}{2} (\ln2 + \ln3)
- \dfrac{1}{2} (\ln3 - \ln2) \)
\( = 3 \ln2 + 2 \ln3 \)
Suy ra:
\(
\begin{cases}
a = 3 \\
b = 2
\end{cases}
\Rightarrow a + b = 5
\)
Chọn: A
Page 16
Lời giải:
Đặt \( u = e^x \Rightarrow du = e^x dx \Rightarrow dx = \dfrac{1}{u} du \)
\( \begin{cases}
x = 0 \Rightarrow u = 1 \\
x = \ln 2 \Rightarrow u = 2
\end{cases} \)
\(4 = \int_{0}^{\ln 2} f(e^x)\, dx = \int_{1}^{2} \dfrac{f(u)}{u}\, du \)
\( 2 = \int_{1}^{2} \dfrac{(2x + 1)}{x} f(x)\, dx = \int_{1}^{2} 2f(x)\, dx + \int_{1}^{2} \dfrac{1}{x} f(x)\, dx \)
\( \Rightarrow 2 = 2 \int_{1}^{2} f(x)\, dx + 4 \)
\( \Rightarrow \int_{1}^{2} f(x)\, dx = -1 \)
Chọn: D
Page 17
Lời giải:
Vì \( (x f(x))' = f(x) + x f'(x) \)
\( \Rightarrow \int_0^1 (x f(x))'\,dx = \int_0^1 f(x)\,dx + \int_0^1 x f'(x)\,dx \)
\( \Rightarrow x f(x) \Big|_0^1 = \int_0^1 f(x)\,dx + 6 \)
\( \Rightarrow 2 = \int_0^1 f(x)\,dx + 6 \Rightarrow \int_0^1 f(x)\,dx = -4 \)
Chọn: C
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Lời giải:
\( 2f(x) + x f\left( \dfrac{1}{x} \right) = x,\ \forall x > 0 \)
Thay \( x \) bởi \( \dfrac{1}{x} \), ta có:
\( 2f\left( \dfrac{1}{x} \right) + \dfrac{1}{x} f(x) = \dfrac{1}{x},\ \forall x > 0 \)
\( \Rightarrow 2x f\left( \dfrac{1}{x} \right) + f(x) = 1,\ \forall x > 0 \)
Tóm lại ta có:
\(
\begin{cases}
2f(x) + x f\left( \dfrac{1}{x} \right) = x \\
f(x) + 2x f\left( \dfrac{1}{x} \right) = 1
\end{cases}
\)
\( \Rightarrow 3f(x) = 2x - 1 \Rightarrow f(x) = \dfrac{2}{3}x - \dfrac{1}{3} \)
\( \int_{\frac{1}{2}}^{2} f(x)\, dx = \int_{\frac{1}{2}}^{2} \left( \dfrac{2}{3}x - \dfrac{1}{3} \right)\, dx
= \left( \dfrac{x^2}{3} - \dfrac{x}{3} \right)\Big|_{\frac{1}{2}}^{2} = \dfrac{3}{4} \)
Chọn: C
Page 19
Lời giải:
Xét \( \int_{0}^{1} x f(3x)\, dx \)
Đặt \( u = 3x \Rightarrow x = \dfrac{u}{3} \Rightarrow dx = \dfrac{1}{3}\, du \)
\( 1 = \int_{0}^{1} x f(3x)\, dx = \int_{0}^{3} \dfrac{u}{3} f(u) \cdot \dfrac{1}{3}\, du = \dfrac{1}{9} \int_{0}^{3} u f(u)\, du \)
Suy ra: \( \int_{0}^{3} u f(u)\, du = 9 hay \int_{0}^{3} x f(xx)\, dx = 9 \)
Đặt \( \begin{cases}
u = f(x) \Rightarrow du = f'(x)\, dx \\
dv = x\, dx \Rightarrow v = \dfrac{x^2}{2}
\end{cases} \)
\( 9 = \int_{0}^{3} f(x) \cdot x\, dx = \dfrac{x^2}{2} f(x) \Big|_{0}^{3} - \int_{0}^{3} \dfrac{x^2}{2} f'(x)\, dx \)
\( \Leftrightarrow 9 = \dfrac{9}{2} - \dfrac{1}{2} \int_{0}^{3} x^2 f'(x)\, dx \)
Suy ra \( \int_{0}^{3} x^2 f'(x)\, dx = -9 \)
Chọn: D
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