Find the area under the given curve over the indicated interval.

– $\int_{1}^{6} 2 x \,dx$

The main objective of this question is to find the area of the curve over the indicated interval.

This question uses the concept of the area under the curve. The area under the curve can be calculated by evaluating the integral over the given interval.

We have to find the area of the curve over the given interval.

The interval given is:

$\space x \space = \space 1 \space to \space x \space = \space 6$

So:

$\space y \space = \space 2 x \space and x \space = \space 1 \space to \space 6$

$\space F(x) \space = \space \int_{1}^{6} y \,dy$

We know that:

$\space y \space = \space 2 x$

By putting values, we get:

$\space F(x) \space = \space \int_{1}^{6}2 x \,dx$

$\space F(x) \space = \space 2 \space \int_{1}^{6} x \,dx$

$\space F(x) \space = \space 2 \space \left[ \frac{ x^2 }{ 2 } \right]_{1}^{6}$

By simplifying, we get:

$\space = \space 36 \space – \space 1$

$\space = \space 35$

Thus:

$\space Area \space = \space 35 \space units \space squared$

The area under the given interval is:

$\space Area \space = \space 35 \space units \space squared$

Example

Find the area under the given interval for the two expressions.

•  $\int_{- 1}^{ 1} x^2 \,dx$
•  $\int_{- 1}^{ 1} x^3 \,dx$

We have to find the area of the curve over the given interval.

The interval given is:

$\space x \space = \space – 1 \space to \space x \space = \space 1$

So:

$\space y \space = \space x^2 \space and x \space = \space – 1 \space to \space 1$

$\space F(x) \space = \space \int_{ – 1}^{ 1 } y \,dy$

We know that:

$\space y \space = \space x^2$

By putting values, we get:

$\space F(x) \space = \space \int_{- 1}^{ 1 } x^2 \,dx$

$\space F(x) \space = \space \left[ \frac{ x^3 }{ 3 } \right]_{ – 1 }^{ 1}$

By simplifying, we get:

$\space = \space \frac{2}{3}$

$\space = \space 0. 6 6 6$

Thus:

$\space Area \space = \space 0. 6 6 6 \space units \space squared$

Now for the second expression. We have to find the area of the curve over the given interval.

The interval given is:

$\space x \space = \space – 1 \space to \space x \space = \space 1$

So:

$\space y \space = \space x^3 \space and x \space = \space – 1 \space to \space 1$

$\space F(x) \space = \space \int_{ – 1}^{ 1 } y \,dy$

We know that:

$\space y \space = \space x^3$

By putting values, we get:

$\space F(x) \space = \space \int_{- 1}^{ 1 } x^3 \,dx$

$\space F(x) \space = \space \left[ \frac{ x^4 }{ 4 } \right]_{ – 1 }^{ 1}$

By simplifying, we get:

$\space = \space 0$

Thus:

$\space Area \space = \space 0 \space units \space squared$