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$$\int_1^4 f(x) dx = 5; \int_0^1 f(3x+1) dx = ?$$

I don't know how to do this. Any hint will be helpful.

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    Hint: Substitution – 2017-02-11

2 Answers 2

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Hint:

We want to compute,

$$\int_{0}^{1} f(3u+1) du$$

Let $x=3u+1$. Then $dx=3du$ and continue on.

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    I think it's better to say let $u = 3x+1$ since the other way could be confusing for a new learner. – 2017-02-11
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    This is a good start to the solution... it's a shame that the OP didn't include any work/thoughts/etc... – 2017-02-11
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    I am absolute newbie in integration :( – 2017-02-11
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Let's allow the letter $u$ to stand for the function

$$u(x)=3x+1$$

that is composed with $f(x)$ in the second integral: $(f \circ u)(x)=f(u(x))=f(3x+1).$

Notice also that

$$\frac{du}{dx} = \frac{d}{dx}(3x+1) = \frac{d}{dx}(3x)+\frac{d}{dx}(1) = 3+ 0 = 3.$$

Multiplying both sides of $\frac{du}{dx}=3$ by $dx$ gets

$$du = 3 dx$$

which means that

$$\frac{1}{3} du = dx.$$

Now, we can substitute expressions in terms of $u$ for expressions in terms of $x$ in $\int_0^1 f(3x+1)dx.$ Since we are changing variables, we must change the limits of integration; the integral is evaluated from $0 \leq x \leq 1,$ and since $u=3x+1$ is a function that is continuous and increasing, then substituting our expressions of $x$ in terms of $u$ above forces us to change the limits of integration to $u(0)=3*0+1=1$ and $u(1)=3*1+1=4.$

Now, we change the variable in $\int_0^1f(3x+1)dx$ to $u$ as outlined above. Recall we substitute $3x+1$ as $u$ and $dx$ as $\frac{1}{3}du$ and then change the limits of integration accordingly:

$$\int_0^1 f(3x+1)dx=\int_1^4 f(u)*\frac{1}{3}*du=\frac{1}{3}\int_1^4f(u)du=\frac{1}{3}*5=\frac{5}{3}.$$

The above evaluation uses the fact that $\int_1^4 f(x)dx=5$ given in the original problem; that the variable is $u$ instead of $x$ makes no difference.