Please help with this question?

Question

Evalute the integral.

dx/49+4x^2 upper limit 3, lower limit 1

Answer 1

divide top and bottom by 49, you get integral of [(1/49)dx]/[1+(2x/7)^2] = 1/49 integral of dx/(1+(2x/7)^2), which is arctangent:

1/49 * (7/2) arctan(2x/7) = 1/14 arctan(2x/7) evaluated at 3 and 1.
The answer is 1/14 (arctan (6/7) - arctan (2/7)) = .0307, or (1/14)arctan(28/61) if you want to be exact.

Answer 2

Here is a solution that has all the steps. Notice that the other answerer ironduke skipped so may steps he missed a factor of 2 (although he may correct his answer when he sees that I have called him out on his mistake).

Here's a picture:
http://math.colgate.edu/~kellen/interspace/integral.gif

∫ dx / (49+4x²)

(1/49) ∫ dx / (1 + 4x²/49)

u = 2x/7
dx = 7/2 (du)

convert limits:
x = 1 → u = 2/7
x = 3 → u = 6/7

(1/14) ∫ du / (1+u²)

(1/14) arctan(x)

[ arctan(6/7) - arctan(2/7) ] / 14

That's it. You can simplify it if you want, using a trig identity, to get the following:
arctan(3416 / 2937) / 28

but that's not necessary, and doesn't really simplify it much.

It is approximately 0.0307376152
See: http://www.google.com/search?q=arctan%283416+%2F+2937%29+%2F+28

Answer 3

dx/49+4x^2 upper limit 3, lower limit 1
= 1/7arctan(2x/7) evaluated from 1 to 3
= (1/7)arctan(6/7) - (1/7)arctan(2/7)
= approximately 0.0614752304

Answer 4

Hello

So we have: (1/49)dx + 4x^2), [1,3]

The integral would be
x/49 + (4/3)x^3, [1,3]

Thus we have 3/49 + (4/3) (27) - [(1/49) + (4/3)] = 36.0614 - 1.3537 = 34.7077

You could also use a TI calculator and do
fnInt(1/49+4x^2, x, 1, 3) = 34.707.

Hope this helps

Answer 5

4/3*x^3 +x/49
=(4*9)+3/49-(4/3+1/49)=34.70748299319727891156462585034

Answer 6

∫ dx/(7^2 + (2x)^2)

We need to replace 7^2 with 1, so we make this substitution,
Let, 2x = 7u
2dx = 7du
dx = 7/2 du

∫ 7/2[ du/7^2 + 7^2(u^2)]
∫ 1/14[du/(1 + u^2)]
= 1/14 [tan^-1(u)]
= 1/14 [tan^-1(2x/7)]
= 1/14 [tan^-1(6/7) - tan^-1(2/7)]
= 0.0307