# Definite integral with one limit given

#### Roisincleary

##### New member
How do I do questions like this? (see picture)

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#### MarkFL

##### Super Moderator
Staff member
Hello, and welcome to FMH!

Let's examine the first one:

$$\displaystyle I=\int_0^{\alpha} x^4+x^2\,dx$$

Now, using the FTOC, we may state:

$$\displaystyle I=\left[\frac{x^5}{5}+\frac{x^3}{3}\right]_0^{\alpha}=\frac{\alpha^3}{15}\left(3\alpha^2+5\right)$$

This is equivalent to choice (b). Can you now try the second problem?

#### MarkFL

##### Super Moderator
Staff member
2.) $$\displaystyle I=\int_0^{\alpha} x^{\frac{1}{3}}-x^{-\frac{1}{3}}\,dx=\left[\frac{3}{4}x^{\frac{4}{3}}-\frac{3}{2}x^{\frac{2}{3}}\right]_0^{\alpha}=\frac{3\alpha^{\frac{2}{3}}}{4}\left(\alpha^{\frac{2}{3}}-2\right)$$
3.) $$\displaystyle I=\int_1^{\alpha}\frac{1}{4x-1}\,dx=\left[\frac{1}{4}\ln(4x-1)\right]_1^{\alpha}=\frac{1}{4}\ln\left(\frac{4\alpha-1}{3}\right)$$