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...