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문제별 결과
1
MCQ
오답
If \(v(t) = 2 t - 4\) and at \(t = 0\) position is 4, then \(x(t) =\)
A
\(t^2 - 4 t\)
B
\(t^2 - 4 t - 4\)
\(t^2 - 4 t + 4\)
정답
D
\(2 t^2 - 4 t\)
E
\(2 t^2 - 4 t + 4\)
해설
\(x = t^2 - 4 t + C\), \(x(0) = C = 4\).
2
MCQ
오답
The area between \(y = 4 x^3 + 2\) and the x-axis from \(x = 1\) to \(x = 2\) is
A
\(36\)
B
\(23\)
C
\(20\)
\(17\)
정답
E
\(9\)
해설
\(\displaystyle\int_{1}^{2} (4x^3 + 2) dx = [x^4 + 2x]_1^2 = (16+4)-(1+2) = 17\).
3
MCQ
오답
\(\displaystyle\int_{0}^{1} \sqrt{x}(x + 1) dx =\)
A
\(0\)
B
\(1\)
\(\dfrac{16}{15}\)
정답
D
\(\dfrac{7}{5}\)
E
\(2\)
해설
\(\displaystyle\int_{0}^{1} \left(x^{\dfrac{3}{2}} + x^{\dfrac{1}{2}}\right) dx = \dfrac{2}{5} + \dfrac{2}{3} = \dfrac{16}{15}\).
4
MCQ
오답
Area enclosed by \(y = x^2\) and \(y = x\)
\(\dfrac{1}{6}\)
정답
B
\(\dfrac{1}{3}\)
C
\(\dfrac{1}{2}\)
D
\(\dfrac{5}{6}\)
E
\(1\)
해설
\(\displaystyle\int_{0}^{1} (x - x^2) dx = \dfrac{1}{2} - \dfrac{1}{3} = \dfrac{1}{6}\).
5
MCQ
오답
\(\int \dfrac{d x}{\sqrt{25 - x^2}} =\)
\(\arcsin\left(\dfrac{x}{5}\right) + C\)
정답
B
\(\arcsin x + C\)
C
\(\left(\dfrac{1}{5}\right) \arcsin\left(\dfrac{x}{5}\right) + C\)
D
\(\sqrt{25 - x^2} + C\)
E
\(2 \sqrt{25 - x^2} + C\)
해설
Standard: \(\int dx/\sqrt{a^2 - x^2} = \arcsin\left(\dfrac{x}{a}\right)\).