Find the volume of the solid whose base is the region in the xy-plane bounded by the given curves and whose cross-sections perpendicular to the x-axis are (a) squares, (b) semicircles, and (c) equilateral triangles. for y=x^2,
Find the indefinite integral of (1-sqrtx)/(1+sqrtx)dx I just worked completely through this problem using substitution (u=1+sqrtx) and came to the answer 3+4sqrtx+x+C. This is incorrect. Not sure where I messed up.
Evaluate the integral of 1/(1+sqrtx)^4 from [0,1] I started off letting u=1+sqrtx and du=1/2sqrtx, but then there is no 1/2sqrtx in the equation. What am I suppose to do next? The answer in the book is 1/6.
I don't know if I did these problems correctly. Can you check them? Use Integration by parts to solve problems. integral x^3(lnx)dx u=lnx dv=x^3dx du=1/x v=x^4/4 Answer:(x^3)(lnx)-(x^4/16) integral xcosxdx x cosx 1 sinx 0 -cosx
find integral using table of integrals ) integral sin^4xdx this the formula i used integral sin^n xdx =-1/n sin^n-1xcosx +n-1/n integral sin^n-2 using the formula this is what i got: integral sin^4xdx=-1/4sin^3xcosx+3/4 integral
y = tan(sqrtx) Find dy/dx. So I found it and I got the answer as sec^2(sqrtx) but the answer is (sec^2(sqrtx))/2(sqrtx)...why? ---------------------------------- 2) Find the line which passes through the point (0, 1/4) and is
That's the same as the integral of sin^2 x dx. Use integration by parts. Let sin x = u and sin x dx = dv v = -cos x du = cos x dx The integral is u v - integral of v du = -sinx cosx + integral of cos^2 dx which can be rewritten
Please help me simplify: 4sqrtx^12/24sqrtx^2 *Is this correct: 2sqrt4/12x The answer you get depends upon where you put the parentheses, and you don't show any. I don't see any way of coming up with your answer. 4 (sqrtx)^12/ 24
how do you determine the convergence of : definite integral from 1--> infinity of lnx/(x^3)? i set the problem as lim (R--->infinity) of the integral of lnx/(x^3) from 1--->R, but i can't compute the integral.