According to Shakespeare, who said "a rose by any other name," all of the following are the same.

Where the differentials MUST match the base for the exponent. Just like the derivative of is not the same as the derivative of , the integral is not handled the same either. The usual form of the integral is:
where du must match u.

Sometimes it already matches.

(There should be a 1/4 in front of the integral!)

Sometimes you can make it match with an appropriate "fudge technique".

Note: To make the du match, you can only fix a constant -- not a variable. Constants factor out of integrals. Variables do NOT!

Look back at the integral to see if du is there. This time the (3) is missing. 3 is a constant which we can fix. Multiply the integral by 1 so that we do not change it.

Note: Be sure to keep the reciprocal of the constant in front. With practice, you can omit the in-between steps of changing to u and back.

(The 2x^4 - 7 should be 2x^4 + 7 throughout the problem)

Use the following procedure for integrals involving something raised to a power:

  1. Pick out n and u.
  2. Find du.
  3. Check to see if du is already in the integral.
  4. If the integral is only missing a constant factor for du, put that constant inside the integral and put its reciprocal outside. Now integrate the inside, keeping the constant on the outside.
  5. If the integral needs more than a constant factor for du, try something else (rewrite, factor, multiply out, etc.). Variables do not factor out of integrals.

Problem 1:

Problem 2:

Problem 3:

Even a derivative then y is the antiderivative or y = 6x dx.

is a family of curves, some of which are shown on the graph at right for values of C,
C = -2, 0, 2, and 4.
The figure on the right with a tangent line to each parabola when x = 1, shows that each curve has the same slope (same derivative) for x = 1. The same applies for any other x.

If we are also given additional information that says the curve passes through the point (2,10), then that lets us select the curve from this family that goes through that point.

Since (2,10) is on the graph, x = 2, y = 10 will satisfy the equation. Substitute in those values for x and y then solve for C.

y = 3x² + C
10 = 3 (2)² + C
10 = 12 + C
-2 = C

Therefore: y = 3x² - 2 is the one passing through (2,10)

Additional information or a point given on the curve is there to evaluate the C which will no longer be an arbitrary constant.

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Copyright © 1996 by B. Chambers and P. Lowry. All Rights Reserved.