integral from 2 to 4 of 2x^2-3x+1 — step-by-step solution | Explain MAths

integral from 2 to 4 of 2x^2-3x+1

⚠ 2x is read as the product 2*x: for a power, write 2^x

⚠ 3x is read as the product 3*x: for a power, write 3^x

1) Let's compute the integral:

\int(2 x^{2} - 3 x + 1)dx

\int(2 x^{2})dx - \int(3 x)dx + \int(1)dx

ℹ

\int(2 x^{2})dx

=

2 \cdot \frac{x^{3}}{3}

‖

- \int(3 x)dx

=

- 3 \cdot \frac{x^{2}}{2}

‖

\int(1)dx

=

x

2 \cdot \frac{x^{3}}{3} - 3 \cdot \frac{x^{2}}{2} + x

ℹ

2 \cdot \frac{x^{3}}{3}

=

\frac{2 x^{3}}{3}

‖

- 3 \cdot \frac{x^{2}}{2}

=

- \frac{3 x^{2}}{2}

▷

\frac{2 x^{3}}{3} - \frac{3 x^{2}}{2} + x

2) Let's evaluate over the interval

[2, 4]

:

(\frac{2 \cdot 4^{3}}{3} - \frac{3 \cdot 4^{2}}{2} + 4) - (\frac{2 \cdot 2^{3}}{3} - \frac{3 \cdot 2^{2}}{2} + 2)

ℹ

4^{3}

=

64

‖

4^{2}

=

16

‖

- (\frac{2 \cdot 2^{3}}{3} - \frac{3 \cdot 2^{2}}{2} + 2)

=

- \frac{2 \cdot 2^{3}}{3} + \frac{3 \cdot 2^{2}}{2} - 2

\frac{2 \cdot 64}{3}

- \frac{3 \cdot 16}{2}

+ 4

- \frac{2 \cdot 2^{3}}{3}

+ \frac{3 \cdot 2^{2}}{2}

- 2

ℹ

2 \cdot 64

=

128

‖

3 \cdot 16

=

48

‖

2^{3}

=

8

‖

2^{2}

=

4

‖

4 - 2

=

2

\frac{128}{3}

- \frac{48}{2}

+ 2

- \frac{2 \cdot 8}{3}

+ \frac{3 \cdot 4}{2}

ℹ

\frac{128}{3} - \frac{48}{2}

=

\frac{128 \cdot 2 - 48 \cdot 3}{3 \cdot 2}

‖

2 \cdot 8

=

16

‖

3 \cdot 4

=

12

\frac{128 \cdot 2 - 48 \cdot 3}{3 \cdot 2}

+ 2

- \frac{16}{3}

+ \frac{12}{2}

ℹ

128 \cdot 2

=

256

‖

-48 \cdot 3

=

-144

‖

3 \cdot 2

=

6

‖

2 - \frac{16}{3}

=

\frac{3 \cdot 2 - 16}{3}

‖

\frac{12}{2}

=

6

\frac{256 - 144}{6} + \frac{3 \cdot 2 - 16}{3} + 6

ℹ

256 - 144

=

112

‖

3 \cdot 2

=

6

\frac{112}{6} + \frac{6 - 16}{3} + 6

ℹ

\frac{112}{6}

=

\frac{56}{3}

‖

6 - 16

=

-10

\frac{56}{3} - \frac{10}{3} + 6

\frac{56 - 10}{3} + 6

\frac{46}{3} + 6

\frac{46 + 6 \cdot 3}{3}

\frac{46 + 18}{3}

▶Integral:

\frac{64}{3}

3) The numeric value of:

\frac{64}{3}

is:

▷

\approx 21.3333