Question

Evaluate the definite integral

Answer 1

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

Answer 2

... It is same as last one. power rule.

Answer 3

\[\int\limits_{1}^{2}12x^3 dx+\int\limits_{1}^{2}3x dx-\int\limits_{1}^{2}3 dx\]

Answer 4

am i on the right track?

Answer 5

plug that into wolframalpha.com and youll get the answer

Answer 6

\[12 \int^{2}_{1} x^3 \space dx + 3\int^{2}_{1}x \space dx - 3\int^{2}_{1} \space dx\]

Answer 7

i dont know what to do next

Answer 8

Take the integral, then apply fundamental theorem\[\int^{a}_{b} f(x) dx = F(b) - F(a)\]

Answer 9

can you show me step by step

Answer 10

Do you know how to compute \[\int x^3 dx\]

Answer 11

x^3/3 so it will be 4

Answer 12

\[\int x^n dx = \frac{x^{n + 1}}{n + 1} + C\]


So \[\int x^3 dx = \frac{x^{(3 + 1)}}{3 + 1} + C = \frac{x^4}{4} + C\]

Answer 13

okay

Answer 14

what happens next

Answer 15

Can you compute

\[\int x \space dx\]

Answer 16

2/2

Answer 17

@Hero

Answer 18

x^2/2

Answer 19

yes

Answer 20

now whats the next step

Answer 21

evaluate at \(2\) and at \(1\) and subtract
i.e. compute
\[\frac{2^2}{2}-\frac{1^2}{2}\]

Answer 22

that will give you
\[\int_1^2xdx\]

Answer 23

geometry will do it as well, but perhaps that is easiest

Answer 24

im confuse i got
\[12\int\limits_{1}^{2}x^4/4+3\int\limits_{1}^{2} x^2/2\]

Answer 25

yes you are confused
find the "anti derivative" then substitute
here is the solution to one of them
\[3\int_1^2xdx=\left.3\frac{x^2}{2}\right|_1^2=3[\frac{2^2}{2}+\frac{1^2}{1})=3(2-\frac{1}{2}=\frac{9}{4}\]

Answer 26

im confuse

Answer 27

i been trying and trying but i can't

Answer 28

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

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

\[12\int\limits_{1}^{2}x^3dx+3\int\limits_{1}^{2}xdx-3\int\limits_{1}^{2}dx\]

\[(12*\frac{1}{4}*x^4+3*\frac{1}{2}*x^2-3x)|_1^2\]

\[(3*x^4+\frac{3}{2}*x^2-3x)|_1^2\]

\[(3*2^4+\frac{3}{2}*2^2-3*2)-(3+\frac{3}{2}-3)\]

\[\int\limits_{1}^{2}(12x^3+3x-3)dx=46.5\]