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Integral of xa
(Because)• When a≠−1 xadx=+C Among them, when a=0 → x0dx=1dx=x+C kdx=kx+C (k:constant) • When a=−1 x−1dx=dx=log|x|+C • When a≠−1 According to the Differentiation Formulas:xa=axa−1 ()=×(a+1)xa=xa Hence, xadx=+C Especially, when a=0 i.e. when x0=1 (kx)=k → kdx=kx+C • When a=−1, as shown in the right Figure 1, there is no function xa whose differential is x−1 or . (log|x|)=→dx=log|x|+C
• When x>0,
(log(x))= → dx=log x+C • When x<0, (log(−x))=×(−1)=→dx=log(−x)+C • As Figure 2, the graph of y=log|x| is separated into 2 parts. The right hand side is y=log(x) (x>0) and the left hand side is y=log(−x) (x<0). They are called y=log|x| as a whole, but are not continous at x=0. |
Examples (1)3x2dx=3×+C=x3+C (2)x−4dx=+C=−+C (3)dx=xdx=+C=+C=x +C (4)dx=3log|x|+C
Figure 1
Figure 2
 But, if a=0 x0=1=0 Hence, there is no function xa whose differential is x−1 or The integral of x−1 or is found in logarithmic functions.  |
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Question 1Find the integrals.
(Step 1)Select one of the questios from the left column.
(Step 2)Select the integral corresponding to the question. (Repeat) Step 1→Step 2 |
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Sum and Difference Rules
(A→)You can exchange the order of integration and sum(or difference).{ f(x)±g(x) }dx=f(x)dx±g(x)dx …(A) Constant Multiple Rule kf(x)dx=kf(x) …(B) (B→)Also, you can exchange the order of integration and multiplication by constant. Examples
(1)(x2+x)dx
=++C
x2dx=+C, xdx=+C
But, you should not write (x2+x)dx=++2C You should write (x2+x)dx=++C In detail, x2dx=+C1, xdx=+C2 (where C1 and C2 are arbitrary constant [not always the same value].) (x2+x)dx=++C1+C2 When C1 and C2 are arbitrary constants, C=C1+C2 is also an arbitrary constant. Thus, you can write (should write) (x2+x)dx=++C |
(2)5x dx
=5×+C=x2+C
Like Example (1), you have to add only one arbitrary constant C.
(3)(x+1)(x+2)dx
(A)(B) show the exchangeability of sum and integralation, multiplication by constant and integration. About product of functions the following is not true.f(x)g(x)dx=f(x)dxg(x)dx When integrands (functions to be integrated) are given by product of functions, first you have to expand the brackets and make them the sum of functions. (x+1)(x+2)dx=(x2+3x+2)dx=+x2+2x+C
(4)(x+1)2dx
Expand the integrand (functions to be integrated) first.(x+1)2dx=(x2+2x+1)dx=+x2+x+C
(5)dx
Separate the fractions first.dx=(−)dx=(2x−2−x−3)dx =2×−+C=−++C |
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Question 2Find the integrals.
(Step 1)Select one of the questios from the left column.
(Step 2)Select the integral corresponding to the question. (Repeat) Step 1→Step 2 |
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