Question

4.用牛顿-莱布尼茨公式求定积分+_0^1(2x+5)dx

Answer 1

问：4.用牛顿-莱布尼茨公式求定积分+_0^1(2x+5)dx

答：①原积分=∫(0到π)√[(sinx)^2-2sinxcosx+(cosx)^2]dx=∫(0到π)√(sinx-cosx)^2dx=∫(0到π/4)(cosx-sinx)dx+∫(π/4到π)(sinx-cosx)dx=(sinx+cosx)(x=π/4)-(sinx+cosx)(x=0)+(﹣cosx－sinx)(x=π)－(﹣cosx－sinx)(x=π/4)=2√2.②原积分=∫(-2到-1)x^4dx+∫(-1到1)dx+∫(1到3)x^4dx=(1/5×x^5)(x=-1)-(1/5×x^5)(x=-2)+x(x=1)-x(x=-1)+(1/5×x^5)(x=3)-(1/5×x^5)(x=1)=283/5.