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#### 2.7.51 Algebra Tutorial Exercise 4

The hard part is that V R + is no longer sufficient to add up all the contributions from the slices, since the slices have varying coefficients. So first we must come up with a vector of these coefficients. Here’s one way:

```2:  -1                 2:  3                    1:  [4, 2, ..., 4]
1:  [1, 2, ..., 9]     1:  [-1, 1, ..., -1]         .
.                      .

1 n v x 9 RET          V M ^  3 TAB             -

```
```1:  [4, 2, ..., 4, 1]      1:  [1, 4, 2, ..., 4, 1]
.                          .

1 |                        1 TAB |
```

Now we compute the function values. Note that for this method we need eleven values, including both endpoints of the desired interval.

```2:  [1, 4, 2, ..., 4, 1]
1:  [1, 1.1, 1.2,  ...  , 1.8, 1.9, 2.]
.

11 RET 1 RET .1 RET  C-u v x

```
```2:  [1, 4, 2, ..., 4, 1]
1:  [0., 0.084941, 0.16993, ... ]
.

' sin(x) ln(x) RET   m r  p 5 RET   V M \$ RET
```

Once again this calls for V M * V R +; a simple * does the same thing.

```1:  11.22      1:  1.122      1:  0.374
.              .              .

*              .1 *           3 /
```

Wow! That’s even better than the result from the Taylor series method.