\(已知L为抛物线y^2=x上从点A(1,-1)到点B(1,1)的一段弧,则\int_{L}xyds=(\,)\)
A: \[\frac{4}{5}\]
B: \[\frac{3}{5}\]
C: \[\frac{2}{5}\]
D: \[\frac{1}{5}\]
A: \[\frac{4}{5}\]
B: \[\frac{3}{5}\]
C: \[\frac{2}{5}\]
D: \[\frac{1}{5}\]
举一反三
- \(已知L是抛物线y=x^2上点O(0,0)与点A(1,1)之间的一段弧,则\int_{L}\sqrt{y}ds=(\,)\) A: \[\frac{1}{12}(5\sqrt{5}-1)\] B: \[\frac{1}{12}(3\sqrt{3}-1)\] C: \[\frac{1}{13}(5\sqrt{5}-1)\] D: \[\frac{1}{13}(3\sqrt{3}-1)\]
- \(已知曲线弧L:y=\sqrt{1-x^2}(0\le x\le 1).则\int_{L}xyds=(\,)\) A: \[1\] B: \[\frac{1}{2}\] C: \[\frac{1}{3}\] D: \[\frac{1}{4}\]
- 已知\(L\)为抛物线\({y^2} = x\) 上从点\(A\left( {1, - 1} \right)\) 到点\(B\left( {1,1} \right)\) 的一段弧,则\(\int_{\;L} {xyds} {\rm{ = }}\)( )。 A: \({3 \over 5}\) B: \({4 \over 3}\) C: \({5 \over 3}\) D: \({4 \over 5}\)
- 微分方程\(2y''+5y'=5x^2-2x-1\)的通解是( )。 A: \(y=C_1+C_2e^{-\frac{5}{2}x}+\frac{1}{3}x^3-\frac{3}{5}x^2+\frac{7}{25}x\) B: \(y=C_1+C_2e^{-\frac{5}{2}x}+\frac{1}{3}x^3-\frac{3}{5}x^2\) C: \(y=C_1+C_2e^{-\frac{5}{2}x}+\frac{1}{3}x^3+\frac{7}{25}x\) D: \(y=C_1+C_2e^{-\frac{5}{2}x}-\frac{3}{5}x^2+\frac{7}{25}x\)
- \(\int { { {\sin }^{2}}x { { \cos }^{5}}xdx}\)=( ) A: \(\frac{1}{3} { { \sin }^{3}}x-\frac{2}{5} { { \sin }^{5}}x+\frac{1}{7} { { \sin }^{7}}x+C\) B: \(\frac{2}{3} { { \sin }^{3}}x-\frac{1}{5} { { \sin }^{5}}x-\frac{1}{7} { { \sin }^{7}}x+C\) C: \(\frac{1}{3} { { \cos }^{3}}x-\frac{2}{5} { { \cos }^{5}}x+\frac{1}{7} { { \cos }^{7}}x+C\) D: \(\frac{2}{3} { { \cos }^{3}}x-\frac{1}{5} { { \cos }^{5}}x-\frac{1}{7} { { \cos }^{7}}x+C\)