Use form of the definition of the integral given in theorem 4 to evaluate integral

Theorem 4 says that if f is integrable on [a,b], then

$\displaystyle \int_{a}^{b} f(x)dx=\lim_{n \rightarrow \infty}\sum_{i=1}^nf(x_i)\Delta{x}$

where $\displaystyle \Delta{x}=\frac{b-a}{n}$ and $\displaystyle x_i=a+i\Delta{x}$

My question is $\displaystyle \int_{-1}^{5} (1+3x)dx$

I have $\displaystyle \Delta{x}=\frac{6}{n}$ ; $\displaystyle x_i=\frac{6i}{n}$

so $\displaystyle \int_{-1}^{5} (1+3x)dx=\lim_{n \rightarrow \infty}\sum_{i=1}^n\left[1+\frac{18i}{n}\right]\cdot\frac{6}{n}=$

$\displaystyle =\lim_{n \rightarrow \infty}\frac{6}{n}\left[\sum_{i=1}^n1+\frac{18}{n}\sum_{i=1}^ni\right]=$

$\displaystyle =\lim_{n \rightarrow \infty}\frac{6}{n}\left[n+\frac{18}{n}\cdot\frac{n(n+1)}{2}\right]=$

$\displaystyle =\lim_{n \rightarrow \infty}6+54\cdot\frac{n^2+n}{n^2}=60$

but by taking the antiderivative I get 42.

Re: Use form of the definition of the integral given in theorem 4 to evaluate integra

Quote:

Originally Posted by

**amthomasjr** Theorem 4 says that if f is integrable on [a,b], then

$\displaystyle \int_{a}^{b} f(x)dx=\lim_{n \rightarrow \infty}\sum_{i=1}^nf(x_i)\Delta{x}$

where $\displaystyle \Delta{x}=\frac{b-a}{n}$ and $\displaystyle x_i=a+i\Delta{x}$

My question is $\displaystyle \int_{-1}^{5} (1+3x)dx$

I have $\displaystyle \Delta{x}=\frac{6}{n}$ ; $\displaystyle x_i=\frac{6i}{n}$

Your $\displaystyle x_i$ is incorrect, it should be $\displaystyle x_i=-1+i\Delta x$

Re: Use form of the definition of the integral given in theorem 4 to evaluate integra

Ah, I see. Forgive me for wasting your time.

Re: Use form of the definition of the integral given in theorem 4 to evaluate integra

Quote:

Originally Posted by

**amthomasjr** Ah, I see. Forgive me for wasting your time.

You learn by asking. Sounds trite, but it's true. You wasted no one's time.

-Dan

Re: Use form of the definition of the integral given in theorem 4 to evaluate integra

Quote:

Originally Posted by

**Plato** Your $\displaystyle x_i$ is incorrect, it should be $\displaystyle x_i=-1+i\Delta x$

$\displaystyle =\lim_{n \rightarrow \infty}\frac{6}{n}\left[\sum_{i=1}^n-2+\frac{18}{n}\sum_{i=1}^ni\right]=$

Instead of getting 6 + 54, I get -12 + 54.

Thanks

Re: Use form of the definition of the integral given in theorem 4 to evaluate integra

Quote:

Originally Posted by

**amthomasjr** I get -12 + 54.

But $\displaystyle -12+54=42$ which is correct.

Re: Use form of the definition of the integral given in theorem 4 to evaluate integra

Yes, that is what I meant.

Thank you