Question

(a) use integration by parts to show that \(\int {f\left( x \right)} dx = xf\left( x \right) - \int {x{f^'}\left( x \right)} dx\)

(b) If f and g are inverse function and f’ is continuous, prove that\(\int\limits_a^b {f(x)dx = bf(b) - af(a)} - \int\limits_{f(a)}^{f(b)} {g(y)dy} \)

(c)In the case where f and g are positive functions and \(b > a > 0\),draw a diagram to give a geometric interpretation of part(b)

(d) Use part (b) to evaluate\(\int\limits_1^e {lnx\,dx} \)

Short answer

We can prove the following by using integration by parts:

\(\int {f(x)dx = xf(x) - \int {x{f^'}(x)dx} } \)

Step-by-step solution

Apply the integration by parts

\(\begin{aligned}{l}uv = \int {vdu} + \int {udv} \\\int {udv = uv - \int {vdu\,\,\,\,\,\,\,\,\,\,........\left( 1 \right)} } \end{aligned}\)

Substitute \(u = f(x)\)

\(\begin{aligned}{l}u = f(x)\\du = {f^'}(x)dx\\dv = dx\\v = x\end{aligned}\)

\(\begin{aligned}{l}\int {udv = uv - \int {vdu\,\,} } \\\int {f(x)dx} = xf(x) - \int {x{f^'}(x)dx} \end{aligned}\)

Therefore,we can prove the following by using integration by parts:

\(\int {f(x)dx = xf(x) - \int {x{f^'}(x)dx} } \)

Answer (b):

We can prove the following by using the part (a)

\(\int\limits_a^b {f(x)dx = bf(b) - af(a)} - \int\limits_{f(a)}^{f(b)} {g(y)dy} \)

Use \({f^'}(x) = g(y)\)

\(\begin{aligned}{l}\int\limits_a^b {f(x)dx} = xf(x) - \int\limits_a^b {{f^'}(x)dx} \\\int\limits_a^b {f(x)dx} = \left( {xf(x)} \right)_a^b - \int\limits_{f(a)}^{f(b)} {g\left( y \right)dy} \end{aligned}\)

Apply the result of part (a)

\(\begin{aligned}{l}\int\limits_a^b {f(x)dx} = \left( {xf(x)} \right)_a^b - \int\limits_{f(a)}^{f(b)} {g\left( y \right)dy} \\\int\limits_a^b {f(x)dx} = af(a) - bf(b) - \int\limits_{f(a)}^{f(b)} {g\left( y \right)dy} \end{aligned}\)

Hence, we can prove that by using the result of part (a)

\(\int\limits_a^b {f(x)dx} = xf(x) - \int\limits_a^b {{f^'}(x)dx} \)

Answer (c):

The graph of part (b) can be given below:

Draw the diagram

In case where \(b > a > 0,f,g > 0\) we have the following diagram for part (b)

Hence, the diagram can be given as below:

Answer (d):

The solution of the given integral by using part(b) is given below:

\(\int\limits_1^e {\ln x\,dx} = 1\)

Need to find

We need to find\(\int\limits_1^e {\ln x\,dx} \)by using the part (b)

From part (b) we can write,

\(\int\limits_a^b {f(x)dx} = xf(x) - \int\limits_a^b {{f^'}(x)dx} \)

Evaluate the integral

\(\int\limits_1^e {\ln x\,dx} = e\ln e - 1\ln 1 - \int\limits_0^1 {{e^y}\,dy} \)

\({e^y}\)and \(x\) are inverse function

\(\begin{aligned}{l}\int\limits_1^e {\ln x\,dx = e - e + 1} \\\int\limits_1^e {\ln x\,dx = 1} \end{aligned}\)

Hence, we can evaluate the given integral by using part (b):

\(\int\limits_1^e {\ln x\,dx = 1} \).