Answer to Question #234648 in Calculus for Sayem

Question #234648

Find the derivative of the function.

  1. Integrate[(Sqrt[t+Sin[t]]),{t,x,1}]
  2. Integrate[Sin[Power[t,4]],{t,2x,3x+4}]
1
Expert's answer
2022-01-31T16:19:34-0500

1.

F(z)=0zt+sintdtF(z)=\int^z_0 \sqrt{t+sint}dt


F(z)=z+sinzF'(z)= \sqrt{z+sinz}


x1t+sintdt=01t+sintdt0xt+sintdt=F(1)F(x)\int^1_x \sqrt{t+sint}dt=\int^1_0 \sqrt{t+sint}dt-\int^x_0 \sqrt{t+sint}dt=F(1)-F(x)


ddx(F(1)F(x))=F(1)F(x)=1+sin1x+sinx\frac{d}{dx}(F(1)-F(x))=F'(1)-F'(x)=\sqrt{1+sin1}-\sqrt{x+sinx}


2.

F(z)=0zsin4tdtF(z)=\int^z_0 sin^4 tdt


F(z)=sin4zF'(z)= sin^4 z


2x3x+4sin4tdt=03x+4sin4tdt02xsin4tdt=F(3x+4)F(2x)\int^{3x+4}_{2x} sin^4 tdt=\int^{3x+4}_{0} sin^4 tdt-\int^{2x}_{0} sin^4 tdt=F(3x+4)-F(2x)


ddx(F(3x+4)F(2x))=3F(3x+4)2F(2x)=\frac{d}{dx}(F(3x+4)-F(2x))=3F'(3x+4)-2F'(2x)=


=3sin4(3x+4)2sin4(2x)=3sin^4(3x+4)-2sin^4(2x)


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