Integration of Algebraic Functions


The integral of the expression xnx^n where n1n \neq -1 is:

xndx=xn+1n+1+C \int x^ndx = \frac{x^{n+1}}{n+1} + C

The integral of x1x^{-1} is:

1xdx=lnx+C \int \frac{1}{x}\,dx = \ln \left| x \right| + C

Note the absolute value inside the natural log.

The integral of the exponential function ex e^x is:

exdx=ex+C \int e^x \, dx = e^x + C


Find a: 0lnaexdx=100 \displaystyle \int_0^{\ln a} e^x \, dx = 100

Since exdx=ex+C \int e^x \, dx = e^x + C , we see that:

0lnaexdx=100[ex]0lna=100elnae0=100a=101 \begin{aligned} \int_0^{\ln a} e^x \, dx &= 100 \\ \left[ e^x \right]_0^{\ln a} &= 100 \\ e^{\ln a}-e^0 &= 100 \\ a &= 101 _\square \end{aligned}


Evaluate: 017x6+42+2x3dx \displaystyle \int_0^1 7x^6 + 42 + \frac{2}{x^3} \, dx .

We can rewrite this problem so that all terms are in the form xnx^n:

017x6+42x0+2x3dx \displaystyle \int_0^1 7x^6 + 42x^0 + 2x^{-3} \, dx

Now, let's integrate, using the rules above:

017x6+42+2x3dx=017x6dx+0142x0dx+012x3dx=[7(x6+16+1)+42x1+2(x3+13+1)]01=x7+42xx201=(17+42(1)1)(0)=42 \begin{aligned} \int_0^1 7x^6 + 42 + 2x^{-3} \, dx &= \int_0^1 7x^6 dx + \int_0^1 42 x^0 \, dx + \int_0^1 2x^{-3}\, dx \\ &= \left[ 7\left(\frac{x^{6+1}}{6+1} \right ) + 42x^1 + 2 \left( \frac{x^{-3+1}}{-3+1} \right ) \right]_0^1 \\ &=\left. x^7 + 42x -x^{-2} \right|_0^1 \\ &= \left(1^7 +42(1)-1\right)-\left(0\right) \\ &= 42 \, _\square \end{aligned}

Note by Arron Kau
7 years, 4 months ago

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