由曲线y²=x,y=x,所围图形的面积是(?)A.∫(1 0)(x²-x)dx B.∫(1 0)(x-x²)dx C.∫(1 0)(y²-x)dy D.∫(1 0)(y-y²)dy
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![由曲线y²=x,y=x,所围图形的面积是(?)A.∫(1 0)(x²-x)dx B.∫(1 0)(x-x²)dx C.∫(1 0)(y²-x)dy D.∫(1 0)(y-y²)dy](/uploads/image/z/2487267-27-7.jpg?t=%E7%94%B1%E6%9B%B2%E7%BA%BFy%26%23178%3B%3Dx%2Cy%3Dx%2C%E6%89%80%E5%9B%B4%E5%9B%BE%E5%BD%A2%E7%9A%84%E9%9D%A2%E7%A7%AF%E6%98%AF%EF%BC%88%3F%EF%BC%89A.%E2%88%AB%281+0%29%28x%26%23178%3B-x%29dx+B.%E2%88%AB%281+0%29%28x-x%26%23178%3B%29dx+C.%E2%88%AB%281+0%29%28y%26%23178%3B-x%29dy+D.%E2%88%AB%281+0%29%28y-y%26%23178%3B%29dy)
由曲线y²=x,y=x,所围图形的面积是(?)A.∫(1 0)(x²-x)dx B.∫(1 0)(x-x²)dx C.∫(1 0)(y²-x)dy D.∫(1 0)(y-y²)dy
由曲线y²=x,y=x,所围图形的面积是(?)
A.∫(1 0)(x²-x)dx B.∫(1 0)(x-x²)dx C.∫(1 0)(y²-x)dy D.∫(1 0)(y-y²)dy
由曲线y²=x,y=x,所围图形的面积是(?)A.∫(1 0)(x²-x)dx B.∫(1 0)(x-x²)dx C.∫(1 0)(y²-x)dy D.∫(1 0)(y-y²)dy
答案为D,根据积分的意义求解