1. Antiderivative

hi,

Q1. Write the gerneral form of the antiderivative of
f(x) = 4x - 2x^2 + x*squareroot(x) + e^x

b) Hence calculate

f (4x - 2x^2 + x*squareroot(x) + e^x) dx

*where f is an antiderivative, a = 0, b = 1.

thanks,

moon

2. Originally Posted by Moon Hoplite
hi,

Q1. Write the gerneral form of the antiderivative of
f(x) = 4x - 2x^2 + x*squareroot(x) + e^x

b) Hence calculate

f (4x - 2x^2 + x*squareroot(x) + e^x) dx

*where f is an antiderivative, a = 0, b = 1.

thanks,

moon

a) $f(x) = 4x - 2x^2 + x \sqrt{x} + e^x$.

Remember that $x \sqrt{x} = x \times x^{\frac{1}{2}} = x^{\frac{3}{2}}$.

So $F(x) = \int{f(x)\,dx} = \int{4x - 2x^2 + x^{\frac{3}{2}} + e^x\,dx}$

$= 2x^2 - \frac{2}{3}x^3 + \frac{2}{5}x^{\frac{5}{2}} + e^x + C$.

For b) evaluate $F(1) - F(0)$.

3. Originally Posted by Prove It
a) $f(x) = 4x - 2x^2 + x \sqrt{x} + e^x$.

Remember that $x \sqrt{x} = x \times x^{\frac{1}{2}} = x^{\frac{3}{2}}$.

So $F(x) = \int{f(x)\,dx} = \int{4x - 2x^2 + x^{\frac{3}{2}} + e^x\,dx}$

$= 2x^2 - \frac{2}{3}x^3 + \frac{2}{5}x^{\frac{5}{2}} + e^x + C$.

For b) evaluate $F(1) - F(0)$.

so does b) = 3.1516??

4. Originally Posted by Moon Hoplite
so does b) = e??
$F(1) = 2(1)^2 - \frac{2}{3}(1)^3 + \frac{2}{5}(1)^{\frac{5}{2}} + e^1 + C$

$= 2 - \frac{2}{3} + \frac{2}{5} + e + C$

$= \frac{26}{15} + e + C$.

$F(0) = 2(0)^2 - \frac{2}{3}(0)^3 + \frac{2}{5}(0)^{\frac{5}{2}} + e^0 + C$

$= 1 + C$.

So $F(1) - F(0) = \frac{26}{15} + e + C - (1 + C)$

$= \frac{11}{15} + e$.

So no, b) is not e. It's $\frac{11}{15} + e$.