Integrate by parts:
∫ uv dx = u ∫ v dx - ∫ (u' ∫ v dx) dx
Let u = -2x
Let v = cos 3x
→ u' = d/dx -2x = -2
→ ∫ -2x cos 3x dx = -2x ∫ cos 3x dx - ∫ (-2 ∫ cos 3x dx) dx
= -2x/3 sin 3x - ∫ -2/3 sin 3x dx
= -2x/3 sin 3x - 2/9 cos 3x + c
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Let u=-2x and let dv/dx = -cos3xthen du/dx = 2 and v = (sin3x)/3
So, I = uv - int[du/dx*v]dx = 2x*sin(3x)/3 - int[2*sin(3x)/3]dx
= 2x/3*sin(3x) - 2/9*cos(3x) + C
The integral would be 10e(1/10)x+c
Let's call the function f and its integral F. If you evaluate F from point a to point b, you will get the area between f and the x-axis. (The area above the x-axis is positive while the area below is negative.) For example, let's say f = 9x2. F (the integral) would then equal 3x3 + C. If we want to find the area between the f and the x-axis from x = 1 to x = 3, we could F from 1 to 3: 3(3)3 - 3(1)3 = 78, so we know that the area between the x-axis and f from x = 1 to x = 3 is 78 square units. (It's all positive in this case, since it's all above the x-axis.) The C in the integral is a constant. It does not matter when you are finding the area under f. If you were to put a number in for C, you would get a function link 3x3 + 7. The derivative if this function is f, so f is the slope, or rate of change, of it's integral. (It doesn't matter what the constant is, since the derivative of a constant is zero. The function 3x3 - 9, for example, has the same derivative.)
A function--namely a parabola (concave up). To "evaluate" this function you would need an x value and would find the resulting y value. To "solve" this function, you would probably be given a y value and asked to find the corresponding x value(s).
the new problem would be 9(7)-2(7) when you substitute 7 for x.The answer is 63-14= 49 So your answer is 49(:
In this specific example one would need to use the u substitution method. * Set u to be x - 3 * Derive x - 3 * u = x - 3 * du = dx Now that we have integrated u we can remove the x - 3 and substitute in u and remove the dx and substitute in du. This is what we have after substituting: * (the integrand of) tan(u)du Now integrate tan(u)du * the Integral of tan(u)du is: * sec2(u) Now resubstitute what we set as u. In this case we set x - 3 to u. This will give us our final answer and integral of tan(x-3)dx. * sec2(x - 3)