Integration formula

If you remember back to the power rule for differentiation, the derivative of ax^n is:

a\times n\times x^{n-1}

We can rearrange this to find the inverse, which gives our formula for integration:

\int ax^n\space dx = a\div(n+1) \times x^{n+1}

We use c here because we don’t know what the constant term on the end was. This constant was removed during differentiation - we can’t know what it will be after we integrate.

Find \int 4x+3 \space dx

Find the integral of \frac{dy}{dx}=9x-4

Given that f'(x)=\sqrt x-4x^{-\frac34} and that f(0)=3, find f(x)

TODO: finish and correct example

flashcards

QuestionAnswer
What is the formula for the integral of ax^n with respect to x?\int ax^n\space dx = a \div (n+1) \times x^{n+1} + c
Why is a constant c added when calculating an indefinite integral?The constant term was removed during differentiation, so we can’t know what it will be after we integrate.
How do you find \int 4x+3 \space dx?\int 4x\space dx = 4\div(1+1)\times x^{1+1} = 2x^2 and \int 3\space dx = 3\div(0+1)\times x^{0+1} = 3x, so answer is 2x^2+3x+c.
What is the result of integrating \frac{dy}{dx}=9x-4?\int 9x^1\space dx = 9\div(1+1)\times x^{1+1} = \frac{9}{2}x^2 and \int 4x^0\space dx = 4\div(0+1)\times x^{0+1} = 4x, so answer is \frac{9}{2}x^2 - 4x + c.
How do you integrate \sqrt x using the power rule?Rewrite \sqrt x as x^{\frac{1}{2}}, then apply \int x^{\frac{1}{2}} dx = 1\div(\frac{1}{2}+1)\times x^{\frac{1}{2}+1} = \frac{2}{3}x^{\frac{3}{2}}.
Given f'(x)=\sqrt x-4x^{-\frac{3}{4}} and f(0)=3, how do you integrate the second term -4x^{-\frac{3}{4}}?\int 4x^{-\frac{3}{4}} dx = 4\div(-\frac{3}{4}+1)\times x^{-\frac{3}{4}+1} = 4\div\frac{1}{4} \times x^{\frac{1}{4}} = 16x^{\frac{1}{4}}, so the term becomes -16x^{\frac{1}{4}}.