Question

If $I={\int }_a^{b}f\left(g\right(x\left)\right).g^{\prime }\left(x\right)dx$ then on substituting g(x) = t where the equation g(x) = t is continuous in the interval [a, b] , I will be equal to -

• A. $I={\int }_a^{b}f\left(t\right).dt$
• B. $I={\int }_{g\left(a\right)}^{g\left(b\right)}f\left(t\right).dt$
• C. $I={\int }_{g\left(b\right)}^{g\left(a\right)}f\left(t\right).dt$
• D. None of these

Answer 1

The correct option is B $I={\int }_{g\left(a\right)}^{g\left(b\right)}f\left(t\right).dt$

We have seen substitution for solving the integrals before also while studying indefinite integral. Substitution here also will be the same but the only difference here we’ll see is that on substituting, limits of variable also change. The only thing we have to take care while substituting is that the equation of substitution (like here, g(x) = t) is not discontinuous in the interval of limits given.
g(x) = t ⇒ g’(x) dx = dt
So, the integrand will become f(t) .dt
Now let’s see how limit changes. Earlier we had limits of “x”. Now we have “t” as variable. So when x = a we’ll have t = g(a), since g(x) = t and similarly when x = b we’ll have t = g(b).
$So,{\int }_a^{b}f\left(g\right(x\left)\right).g^{\prime }\left(x\right)dx={\int }_{g\left(a\right)}^{g\left(b\right)}f\left(t\right).dt$