题目内容
17.若$\int_0^x{{a^2}da={x^2}}$(x>0),则$\int_1^x{|{a-2}|da=}$1.分析 由题意可知$\frac{1}{3}$x3=x2,(x>0),即可求得x的值,根据分段函数定积分的性质,即可求得答案.
解答 解:由题意可知$\int_0^x{{a^2}da={x^2}}$(x>0),则${∫}_{0}^{x}$a2da=$\frac{1}{3}$a3${丨}_{0}^{x}$=$\frac{1}{3}$x3=x2,
则$\frac{1}{3}$x3=x2,(x>0),解得:x=3,
$\int_1^x{|{a-2}|da=}$${∫}_{1}^{x}$丨a-2丨da=${∫}_{1}^{2}$(2-a)da+${∫}_{2}^{3}$(a-2)da,
=(2a-$\frac{1}{2}$a2)${丨}_{1}^{2}$+($\frac{1}{2}$a2-2a)${丨}_{2}^{3}$,
=(4-2)-(2-$\frac{1}{2}$)+($\frac{9}{2}$-6)-(2-4),
=1,
∴$\int_1^x{|{a-2}|da=}$1,
故答案为:1.
点评 本题考查定积分的运算,分段函数的定积分的性质,考查计算能力,属于中档题.
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