Question

$I={\int }_a^{b}f\left(g\right(x\left)\right).g^{\prime }\left(x\right)dx$. If g(x) = t is continuous in the interval [a, b], then I =

• A. ${\int }_a^{b}f\left(t\right).dt$
• B. ${\int }_{g\left(a\right)}^{g\left(b\right)}f\left(t\right).dt$
• C. ${\int }_{g\left(b\right)}^{g\left(a\right)}f\left(t\right).dt$

Answer 1

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

When 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), Similarly, when x = b we’ll have t = g(b).
$So,{\int }_a^{b}f\left(g\right(x\left)\right).g^{\prime }\left(x\right)dx={\int }_{g\left(a\right)}^{g\left(b\right)}f\left(t\right).dt$