Integration

Iam posting the pic of the question. I solved question (a). Not able to solve q(b)

Iam posting the pic of the question. I solved question (a). Not able to solve q(b)

You know how for a graph of $y=f(x)$, the notation

$\displaystyle \int_a^b f(x) dx$

can be interpreted as the signed area between the curve at the x-axis. Right?

In other words, more specific to your case, it will just be the area you measure from the x-axis up to the curve over the interval $\ln 2 < x < \ln 3$.

The notation

$\displaystyle \int_{f(a)}^{f(b)} y dy$

on the other hand, still can be interpreted to be the area, but it's this time it is measured from the y-axis up to the curve.

Therefore, what you have is the following image, whereby the blue represents your answer to part (a) and red represents the integral in part (b)

So do I need to minus the part (a) value

You know how for a graph of $y=f(x)$, the notation

$\displaystyle \int_a^b f(x) dx$

can be interpreted as the signed area between the curve at the x-axis. Right?

In other words, more specific to your case, it will just be the area you measure from the x-axis up to the curve over the interval $\ln 2 < x < \ln 3$.

The notation

$\displaystyle \int_{f(a)}^{f(b)} y dy$

on the other hand, still can be interpreted to be the area, but it's this time it is measured from the y-axis up to the curve.

Therefore, what you have is the following image, whereby the blue represents your answer to part (a) and red represents the integral in part (b)

So when I try to do I get ln3-ln2 which is ln(3/2)

Are you just guessing? What are you calculating? Be clear, I can't read minds.

Are you just guessing? What are you calculating? Be clear, I can't read minds.

I could do question (a) by applying the concept of integration. I dont understand the b part

Are you just guessing? What are you calculating? Be clear, I can't read minds.

I explained what it represents visually.

Rather than integrating directly, you should use the concept I have explained to you.

Look, the red + blue areas make up an L shape overall. You can easily determine the area of an L shape -- you do this at GCSE level.

So figure that out, and if you subtract the blue area from this, you get the red area, hence the integral.

You know how for a graph of $y=f(x)$, the notation

$\displaystyle \int_a^b y dx$

can be interpreted as the signed area between the curve at the x-axis. Right?

In other words, more specific to your case, it will just be the area you measure from the x-axis up to the curve over the interval $\ln 2 < x < \ln 3$.

The notation

$\displaystyle \int_{f(a)}^{f(b)} x dy$

where $x = f^{-1}(y)$ on the other hand, still can be interpreted to be the area, but it's this time it is measured from the y-axis up to the curve.

Therefore, what you have is the following image, whereby the blue represents your answer to part (a) and red represents the integral in part (b)

I explained what it represents visually.

Rather than integrating directly, you should use the concept I have explained to you.

Look, the red + blue areas make up an L shape overall. You can easily determine the area of an L shape -- you do this at GCSE level.

So figure that out, and if you subtract the blue area from this, you get the red area, hence the integral.

I explained what it represents visually.

Rather than integrating directly, you should use the concept I have explained to you.

Look, the red + blue areas make up an L shape overall. You can easily determine the area of an L shape -- you do this at GCSE level.

So figure that out, and if you subtract the blue area from this, you get the red area, hence the integral.

I tried but not getting it

can you not jus try integration by parts?

I tried but not getting it