Integration

Lesson

When calculating the area between a curve and the $x$`x`-axis, it's very much like calculating the definite integral, but we have to be mindful of one thing: is the area above the $x$`x`-axis, below the $x$`x`-axis, or a little bit of both? This will change how we complete our calculations.

Calculate the area bounded by the curve $y=(x-2)(x-4)$`y`=(`x`−2)(`x`−4) and the $x$`x`-axis, between the $x$`x`-intercepts.

First we need to sketch a graph of our function. Since we're only interested in the area between the $x$`x`-intercepts we can locate those easily for this function in factored form, and then note the function is a minimum.

We notice that our area is completely below the $x$`x`-axis. This will mean that our definite integration will return a negative result, but area can't be negative.

We have two ways to combat this in our calculations: either we can multiply our answer by $-1$−1 to change it to a positive, or we can swap the upper and lower limits on our integral symbol. The first option is probably the most common.

Calculate the area bounded by the curve $y=(x+2)^3-1$`y`=(`x`+2)3−1 and the lines $x=-3$`x`=−3 and $x=0$`x`=0.

Again we need a sketch, and using what we know about graphing cubic functions we get the following graph.

Some of the area is below and some is above, so this time we'll break our work into two sections.

Consider the function $y=2x+3$`y`=2`x`+3.

Plot the function.

Loading Graph...Calculate the exact area bounded by the curve $y=2x+3$

`y`=2`x`+3, $x=1$`x`=1, $x=3$`x`=3 and the $x$`x`-axis.

Find the exact area of the shaded region under the curve $y=6x^2$`y`=6`x`2.

Loading Graph...

Find the exact area of the shaded regions bounded by the curve $y=3\cos x$`y`=3`c``o``s``x`.

Loading Graph...

Choose and apply a variety of differentiation, integration, and antidifferentiation techniques to functions and relations, using both analytical and numerical methods

Apply integration methods in solving problems