The antiderivative of #f(x)=x-7# is quite simple. It just involves two cases of the reverse power rule. First, we write down the problem in math notation:
#intx-7dx#
The sum rule for integrals (antiderivatives) says we can simplify this to:
#intxdx-int7dx#
Furthermore, another rule of integrals says that #intaxdx=aintxdx#; in other words, we can pull constants out of the integral. Since the #7# in the second integral is constant, we can say:
#intxdx-7intdx#
Solving these is easy. The power rule for derivatives says that to find the derivative of #x^a#, multiply #x# by #a# and then decrease #a# by one (such as #d/dx(x^2)=2*x^(2-1)=2x^1=2x#). It makes sense, then, that the antiderivative would be this in reverse: increase #a# by one, then divide by the new power (so the antiderivative of #2x# would be #2x^(1+1)=2x^2=(2x^2)/2=x^2#, which is the original function). Applying this to our problem:
#intxdx=x^2/2+C_1# and #-7intdx=-7x-C_2-># this one is the so-called "perfect integral"
Our solution is therefore:
#x^2/2+C_1-7x-C_2#
And since #C_1-C_2# is just another constant, we can combine them into a general constant #C# to get our final result of:
#x^2/2+7x+C#
