solving intergrals

Question

Let g(x) be a continuous function, and G(X) is an antiderivative of g(x).

You are given that G(-2)= -2, G(2)= 1, ∫g(x)dx= 3 b=2 and a=5, and ∫g'(x)dx=21 b=7, a=3.

Find the value of each of the quantities below. If insufficient information is given to find one of the values, explain why.

a). ∫(g(-x)+4x)dx= b=2, a=-2

b) G(5)=

c). the average value of g(x) on [3,7]

d). ∫( (g'(x)(x^2+1) - g(x)(2x))/((x^2+1)^2))dx b=2, a=-2

Answer 1

a.) -G(-x) + 2x² is an antiderivative of g(-x) + 4x.

(Let's check the above: the derivative of -G(-x) + 2x² is -g(-x)(-1) + 4x = g(-x) + 4x.)

Therefore ∫(g(-x)+4x)dx, with limits of integration from a=-2 up to b=2, is

(-G(-b) + 2b²) - (-G(-a) + 2a²)

and now we substitute a=-2 and b = 2,

= (-G(-2) + 2·2²) - (-G(2) + 2(-2)²)

= (2 + 8) - (-1 + 8)

= 10 - 7

= 3

b.) We're given ∫g(x)dx= 3, with limits of integration a=5 and b=2.

(That's what you wrote. I'm surprised that a, the lower limit of integration, is greater than b, the upper limit of integration. Could this be a typo?)

We're given that an antiderivative of g(x) is G(x).

Therefore 3 = ∫g(x)dx = G(2) - G(5) = 1 - G(5)

In other words, 3 = 1 - G(5)

Therefore G(5) = -2

c.) The area under the curve g(x) from x=3 to x=7 is

∫g(x)dx, with limits of integration a=3 and b=7.

We're given that an antiderivative of g(x) is G(x).

Therefore ∫g(x)dx (with limits of integration a=3 and b=7) = G(7) - G(3).

So the area under the curve from x=3 to x=7 is G(7) - G(3).

So the average value of g(x) from x=3 to x=7 is (G(7) - G(3))/4. (See footnote*)

If we knew the values of G(7) and G(4), we could compute this average (G(7) - G(3))/4, but they didn't tell us:

insufficient information!

*Footnote. To see why this is the average value, suppose that the area under the curve from x=3 to x=7 was 20 square inches. This area is 4 units wide (because 7-4=3). Therefore the average value of g(x) from x=3 to x=7 would be 20/4 = 5: you divide the area by the width to get the average height.

d.) ∫( (g'(x)(x²+1) - g(x)(2x))/((x²+1)²))dx

This looks horrible at first sight.

I hope you have memorized the incantation for the derivative of a quotient (i.e., a fraction): "bottom times derivative of top, minus top times derivative of bottom, over bottom squared". We're very lucky here: the integrand

( (g'(x)(x²+1) - g(x)(2x))/((x²+1)²)) miraculously has this form, if we

let (x² + 1) be "the bottom" of the fraction, and

let g(x) be "the top" of the fraction.

Then the derivative of the top is g'(x), and

the derivative of the bottom is 2x,

and the bottom squared is (x² + 1)²

Thus an antiderivative of the horrible integrand

( (g'(x)(x²+1) - g(x)(2x))/((x²+1)²))

is the simple fraction (g(x)) / (x² + 1).

Let's check our work: the derivative of the fraction (g(x)) / (x² + 1) is

"bottom times derivative of top, minus top times derivative of bottom, over bottom squared",

in other words,

((x² + 1)g'(x) - g(x)(2x)) / ((x² + 1)²),

which is the original horrible integrand we started with.

Therefore

∫( (g'(x)(x²+1) - g(x)(2x))/((x²+1)²))dx, with limits of integration a=-2 and b=2 is

(((g(b)) / (b² + 1)) - (((g(a)) / (a² + 1))

= (((g(2)) / (2²+1)) - (((g(-2)) / ((-2)² + 1))

= (((g(2))/5) - (((g(-2)/5)

If we knew what g(2) and g(-2) were, we could compute this number, but they never told us.

Insufficient information!