题目内容
13.如图,由$y={x^2},x=0,y=\frac{1}{4}$所围成阴影部分面积为( )| A. | $\frac{2}{3}$ | B. | $\frac{1}{4}$ | C. | $\frac{1}{2}$ | D. | $\frac{1}{3}$ |
分析 根据题意,求出积分的上下限,即可得出结论.
解答 解:由$\left\{\begin{array}{l}{y={x}^{2}}\\{y=\frac{1}{4}}\end{array}\right.$,可得x=±$\frac{1}{2}$,$\left\{\begin{array}{l}{y={x}^{2}}\\{x=1}\end{array}\right.$,可得y=1,
∴S=${∫}_{0}^{\frac{1}{2}}(\frac{1}{4}-{x}^{2})dx$+${∫}_{\frac{1}{2}}^{1}({x}^{2}-\frac{1}{4})dx$=($\frac{1}{4}x-\frac{1}{3}{x}^{3}$)${|}_{0}^{\frac{1}{2}}$+($\frac{1}{3}{x}^{3}-\frac{1}{4}x$)${|}_{\frac{1}{2}}^{1}$=$\frac{1}{12}$+$\frac{1}{6}$=$\frac{1}{4}$,
故选:B.
点评 本题考查了定积分,考查了数形结合的数学思想,解答此题的关键是明确微积分基本定理.
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