7.1 Integration by Substitution

[WilburWildcat]

Can someone please explain how I am supposed to use Substitution to solve these problems?

Suppose \(\displaystyle \displaystyle{ \int_0^1 \, }\)\(\displaystyle \, g(t)\, dt\, =\, 3.\) Compute each of the following:

\(\displaystyle \mbox{a) }\,\displaystyle{ \int_0^{0.5} \, }\)\(\displaystyle \, g(2t)\, dt\)

\(\displaystyle \mbox{b) }\,\displaystyle{ \int_0^{1} \, }\)\(\displaystyle \, \left(t\, +\, g(1\, -\, t)\right)\, dt\)

\(\displaystyle \mbox{c) }\,\displaystyle{ \int_1^{1.5} \, }\)\(\displaystyle \, g(3\, -\, 2t)\, dt\)

Not looking for answers but the method used to get the answers, Thank you.

 

[Ishuda]

It looks like you want to get an integral in the form of \(\displaystyle \int_0^1 \,g(t)\, dt\) for each answer. Let's look at an example: Suppose I had

A = \(\displaystyle \displaystyle{ \int_{-0.5}^{-0.4} }\)\(\displaystyle \left[3 t^2 \,+ \,g(10t\, +\, 5)\right]\, dt\)

Well, since the integral of a sum is the sum of the integrals

A = \(\displaystyle \displaystyle{\int_{-0.5}^{-0.4}}\)\(\displaystyle 3 t^2 \,dt\, +\, \)\(\displaystyle \displaystyle{\int_{-0.5}^{-0.4}}\)\(\displaystyle g(10t\, +\, 5) \,dt\)

Now, in that second integral suppose we let u = 10 t + 5. Then when t is -0.5, t is zero and when t is -0.4, u is 1. We also have du = 10 dx or dx = (1/10) du. Putting that all together we have

A = \(\displaystyle \displaystyle{\int_{-0.5}^{-0.4} }\)\(\displaystyle 3 t^2 dt + \left(\dfrac{1}{10}\right)\)\(\displaystyle \displaystyle{\int_0^{1}}\)\(\displaystyle g(u) du\)

We should now be able to calculate the value of A since the second integral in the equation is given as 3:

A = \(\displaystyle \displaystyle{\int_{-0.5}^{-0.4}}\)\(\displaystyle 3 t^2 dt \, + \,0.3\)

Of course you would have to evaluate the first integral.

 

[HallsofIvy]

Essentially, you are told what the integral of g(t) is and you want to find the integral of g of various other functions. If you want to reduce g(f(x)) to g(t), the obvious substitution is t= f(x).