Question

Find the area od the region bounded by the x-axis and the graph of the function
f(x)=12x^2-48
between x=1 and x=3

Answer 1

You would solve the following integral for this question:

\[\int\limits_{1}^{3}(12x^{2}-48)dx\]

Answer 2

im so bad at that can you help me out

Answer 3

step by step

Answer 4

do we do \[12\int\limits_{1}^{3}(x^2 dx)+48\int\limits_{1}^{3} dx\]

Answer 5

No, it should be -48 (not +48) ...

Answer 6

oh yeah

Answer 7

Alright, since this region is bounded by the x-axis and this curve, it means that the actual integral of the region you are intending to find is
\[\int\limits_{1}^{3}(12x^{2}-48)-0dx\]
Note the '-0' which defines this region [since it is bound by the line y=0 (aka the x-axis)].
So from the previous integral, we have:
\[\int\limits_{1}^{3}(12x^{2}-48)dx\] =
(4x^3 - 48x) (1-->3) =
[4(27)-48(3)] - (4-48) =
108-144+44 =
8 units^2

Answer 8

how did you do got[4(27)

Answer 9

We sub. x=3 in the expression (see the expression 1-->3? That's actually referring to the limits from 1 to 3 - Open Study doesn't allow for me to put in these limits in its Equation function, so I've represented it this way. By right, you should represent it as shown in the diagram below:

Answer 10

See if you can read this and it is the same as above:
\[[4x^{3}-48x]_{1}^{3}\]

Answer 11

\[f(x)=12x^2-48\]

we have to integrate this function

\[1\leq x\leq3\]

then

\[\int\limits_1^3(12x^2-48)dx\]

\[\int\limits_1^3(12x^2-48)dx=(4x^3-48x)]_1^3\]


\[4*3^3-48*3-(4*1^3-48*1)\]

\[\boxed{\boxed{\int\limits_1^3(12x^2-48)dx=8}}\]