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

**amthomasjr** · November 27th 2012, 12:13 PM

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

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

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

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

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

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

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

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

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

but by taking the antiderivative I get 42.

**Plato** · November 27th 2012, 12:22 PM

Originally Posted by amthomasjr

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

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

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

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

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

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

**amthomasjr** · November 27th 2012, 12:28 PM

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

**topsquark** · November 27th 2012, 12:33 PM

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

**amthomasjr** · November 27th 2012, 12:34 PM

Originally Posted by Plato

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

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

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

Thanks

**Plato** · November 27th 2012, 12:43 PM

Originally Posted by amthomasjr

I get -12 + 54.

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

**amthomasjr** · November 27th 2012, 12:51 PM

Yes, that is what I meant.
Thank you

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