2 of evaluating definite integrals

[Beeorz]

Homework Statement

1)Evaluate the definite integral using FTC:
$$\int_1^4 \left( \frac{d}{dt} \sqrt{4+3t^4} \right)dt$$


2)Evaluate the definite integral:
$$\int_{-2}^6 f(x)dx$$

f(x)=
{x if x<1}
{1/x if x>=1}

Homework Equations

The Attempt at a Solution

Having trouble getting the anti-derivative it seems...
1)$$\int_1^4 \sqrt{4} + \int_1^4 \sqrt{3t^4}$$

$$2\int_1^4 1 + \sqrt{3} \int_1^4 t^2$$

$$2x + \sqrt{3} \frac {t^3}{3}$$



2)$$\int_{-2}^0 x + \int_1^6 \frac {1}{x}$$

$$\int_{-2}^0 \frac {1}{2}x^2 + \int_1^6 \ln \abs{x}$$



Anyways, I hope someone answers :\...took quite some time learning the tex format.

 

[Beeorz]

I figured them both out!

 

[Defennder]

Homework Statement

1)Evaluate the definite integral using FTC:
$$\int_1^4 \left( \frac{d}{dt} \sqrt{4+3t^4} \right)dt$$


2)Evaluate the definite integral:
$$\int_{-2}^6 f(x)dx$$

f(x)=
{x if x<1}
{1/x if x>=1}

Homework Equations

The Attempt at a Solution

Having trouble getting the anti-derivative it seems...
1)$$\int_1^4 \sqrt{4} + \int_1^4 \sqrt{3t^4}$$

$$2\int_1^4 1 + \sqrt{3} \int_1^4 t^2$$

$$2x + \sqrt{3} \frac {t^3}{3}$$

For one thing, you only need observe that the question asks you to find the anti-derivative of a derivative. Your approach appears correct but a little off. You appear to have improperly split up the square root term. $$\sqrt{a+b} \neq \sqrt{a} + \sqrt{b}$$.

2)$$\int_{-2}^0 x + \int_1^6 \frac {1}{x}$$
$$\int_{-2}^0 \frac {1}{2}x^2 + \int_1^6 \ln \abs{x}$$

You're forgetting the interval for 0<x<1.

EDIT: Two minutes too late...