-e-x + C.
One can use integration by parts to solve this. The answer is (x-1)e^x.
The integral would be 10e(1/10)x+c
The antiderivative of a function which is equal to 0 everywhere is a function equal to 0 everywhere.
The fundamental theorum of calculus states that a definite integral from a to b is equivalent to the antiderivative's expression of b minus the antiderivative expression of a.
-e-x + C.
int(e 3x) = (1/3)e 3x ========
The antiderivative, or indefinite integral, of ex, is ex + C.
The antiderivative of 2x is x2.
One can use integration by parts to solve this. The answer is (x-1)e^x.
Powers of e are simple to integrate. The derivative of eu equals u'eu; inversely, the antiderivative of eu equals eu/u'. Therefore, the antiderivative of e1/-x equals (e1/-x)/{d/dx[1/-x]}. The derivative of 1/-x, which can also be expressed as x-1, equals (-1)x(-1-1) = -x-2 = -1/x2.
The integral would be 10e(1/10)x+c
The antiderivative of a function which is equal to 0 everywhere is a function equal to 0 everywhere.
35x2
Using u-substitution (where u = sinx), you'll find the antiderivative to be 0.5*sin2x + C.
I assume you mean -10x^4? In that case, antiderivative would be to add one to the exponent, then divide by the exponent. So -10x^5, then divide by 5. So the antiderivative is -2x^5.
The antiderivative of 1/x is ln(x) + C. That is, to the natural (base-e) logarithm, you can add any constant, and still have an antiderivative. For example, ln(x) + 5. These are the only antiderivatives; there are no different functions that have the same derivatives. This is valid, in general, for all antiderivatives: if you have one antiderivative of a function, all other antiderivatives are obtained by adding a constant.