# Calculus

## Integration of Functions # Integrals of Vector-Valued Functions

## Solved Problems

Click or tap a problem to see the solution.

### Example 7

Find R (t) if $\mathbf{R}^\prime\left( t \right) = \left\langle {\sin \frac{t}{3},\cos \frac{t}{2}} \right\rangle$ and R (π) = ⟨1/2, 1/2⟩.

### Example 8

Compute the integral $\int\limits_0^1 {\left\langle {\frac{{2t}}{{1 + {t^2}}},\frac{2}{{1 + {t^2}}}} \right\rangle dt}.$

### Example 9

Compute the integral $\int\limits_0^{\frac{\pi }{6}} {\left\langle {2\cos t,\sin 2t} \right\rangle dt}.$

### Example 10

A particle starts moving from the origin with the velocity $\mathbf{v}\left( t \right) = \left\langle {4t,3{t^2} - 1} \right\rangle,$ where $$t$$ is measured in seconds and the components are in meters per second. Determine the displacement $$\left| \mathbf{r} \right|$$ of the particle in $$t = 2$$ seconds.

### Example 11

A particle starts moving from the origin with the acceleration $\mathbf{a}\left( t \right) = \left\langle {6t,4} \right\rangle,$ where $$t$$ is measured in seconds and the components are in meters per seconds squared. Determine the displacement $$\left| \mathbf{r} \right|$$ of the particle in $$t = 1$$ seconds if $$\mathbf{v}\left( 0 \right) = \left\langle {2,2} \right\rangle .$$

### Example 7.

Find $$\mathbf{R}\left( t \right)$$ if $\mathbf{R}^\prime\left( t \right) = \left\langle {\sin \frac{t}{3},\cos \frac{t}{2}} \right\rangle$ and $$\mathbf{R}\left( \pi \right) = \left\langle {\frac{1}{2},\frac{1}{2}} \right\rangle .$$

Solution.

Integrating the vector function yields:

$\mathbf{R}\left( t \right) = \int {\left\langle {\sin \frac{t}{3},\cos \frac{t}{2}} \right\rangle dt} = \left\langle {\int {\sin \frac{t}{3}dt} ,\int {\cos \frac{t}{2}dt} } \right\rangle = \left\langle { - 3\cos \frac{t}{3},2\sin \frac{t}{2}} \right\rangle + \mathbf{C}.$

We find the vector $$\mathbf{C} = \left\langle {{C_1},{C_2}} \right\rangle$$ from the initial condition $$\mathbf{R}\left( \pi \right) = \left\langle {\frac{1}{2},\frac{1}{2}} \right\rangle :$$

$\mathbf{R}\left( \pi \right) = \left\langle { - 3\cos \frac{\pi }{3},2\sin \frac{\pi }{2}} \right\rangle + \mathbf{C} = \left\langle { - 3 \cdot \frac{1}{2},2 \cdot 1} \right\rangle + \mathbf{C} = \left\langle { - \frac{3}{2},2} \right\rangle + \mathbf{C} = \left\langle {\frac{1}{2},\frac{1}{2}} \right\rangle .$

Then

$\mathbf{C} = \left\langle {\frac{1}{2},\frac{1}{2}} \right\rangle - \left\langle { - \frac{3}{2},2} \right\rangle = \left\langle {2, - \frac{3}{2}} \right\rangle .$

$\mathbf{R}\left( t \right) = \left\langle { - 3\cos \frac{t}{3},2\sin \frac{t}{2}} \right\rangle + \left\langle {2, - \frac{3}{2}} \right\rangle .$

### Example 8.

Compute the integral $\int\limits_0^1 {\left\langle {\frac{{2t}}{{1 + {t^2}}},\frac{2}{{1 + {t^2}}}} \right\rangle dt}.$

Solution.

Integrating component-by-component, we can write:

$\int\limits_0^1 {\left\langle {\frac{{2t}}{{1 + {t^2}}},\frac{2}{{1 + {t^2}}}} \right\rangle dt} = \left\langle {\int\limits_0^1 {\frac{{2tdt}}{{1 + {t^2}}}} ,\int\limits_0^1 {\frac{{2dt}}{{1 + {t^2}}}} } \right\rangle .$

We evaluate each of the integrals separately. To find the first integral, we make the substitution $$u = 1 + {t^2},$$ $$du = 2tdt.$$ Then

$\int {\frac{{2tdt}}{{1 + {t^2}}}} = \int {\frac{{du}}{u}} = \ln \left| u \right| = \ln \left( {1 + {t^2}} \right),$

and

$\int\limits_0^1 {\frac{{2tdt}}{{1 + {t^2}}}} = \left. {\ln \left( {1 + {t^2}} \right)} \right|_0^1 = \ln 2 - \ln 0 = \ln 2.$

Calculate the second integral:

$\int\limits_0^1 {\frac{{2dt}}{{1 + {t^2}}}} = \left. {2\arctan t} \right|_0^1 = 2\left( {\arctan 1 - \arctan 0} \right) = 2\left( {\frac{\pi }{4} - 0} \right) = \frac{\pi }{2}.$

Thus, the initial integral is represented by the vector

$\int\limits_0^1 {\left\langle {\frac{{2t}}{{1 + {t^2}}},\frac{2}{{1 + {t^2}}}} \right\rangle dt} = \left\langle {\ln 2,\pi } \right\rangle .$

### Example 9.

Compute the integral $\int\limits_0^{\frac{\pi }{6}} {\left\langle {2\cos t,\sin 2t} \right\rangle dt}.$

Solution.

By integrating componentwise, we obtain:

$\int\limits_0^{\frac{\pi }{6}} {\left\langle {2\cos t,\sin 2t} \right\rangle dt} = \left\langle {\int\limits_0^{\frac{\pi }{6}} {2\cos tdt} ,\int\limits_0^{\frac{\pi }{6}} {\sin 2tdt} } \right\rangle = \left\langle {\left. {2\sin t} \right|_0^{\frac{\pi }{6}}, - \left. {\frac{{\cos 2t}}{2}} \right|_0^{\frac{\pi }{6}}} \right\rangle = \left\langle {2\left( {\frac{1}{2} - 0} \right), - \frac{1}{2}\left( {\frac{1}{2} - 1} \right)} \right\rangle = \left\langle {1,\frac{1}{4}} \right\rangle .$

### Example 10.

A particle starts moving from the origin with the velocity $\mathbf{v}\left( t \right) = \left\langle {4t,3{t^2} - 1} \right\rangle,$ where $$t$$ is measured in seconds and the components are in meters per second. Determine the displacement $$\left| \mathbf{r} \right|$$ of the particle in $$t = 2$$ seconds.

Solution.

Determine the position of the particle in $$t = 2$$ seconds by integrating the velocity vector:

$\mathbf{r}\left( t \right) = \int\limits_0^2 {\mathbf{v}\left( t \right)dt} = \int\limits_0^2 {\left\langle {4t,3{t^2} - 1} \right\rangle dt} = \left\langle {\int\limits_0^2 {4tdt} ,\int\limits_0^2 {\left( {3{t^2} - 1} \right)dt} } \right\rangle = \left\langle {\left. {2{t^2}} \right|_0^2,\left. {{t^3} - t} \right|_0^2} \right\rangle = \left\langle {8,8 - 2} \right\rangle = \left\langle {8,6} \right\rangle .$

Hence, the displacement of the particle is equal

$\left| {\mathbf{r}\left( t \right)} \right| = \sqrt {{8^2} + {6^2}} = 10\,\text{m}.$

### Example 11.

A particle starts moving from the origin with the acceleration $\mathbf{a}\left( t \right) = \left\langle {6t,4} \right\rangle,$ where $$t$$ is measured in seconds and the components are in meters per seconds squared. Determine the displacement $$\left| \mathbf{r} \right|$$ of the particle in $$t = 1$$ seconds if $$\mathbf{v}\left( 0 \right) = \left\langle {2,2} \right\rangle .$$

Solution.

First we integrate the acceleration vector to obtain the velocity vector:

$\mathbf{v}\left( t \right) = \int {\mathbf{a}\left( t \right)dt} = \int {\left\langle {6t,4} \right\rangle dt} = \left\langle {3{t^2} + {k_1},4t + {k_2}} \right\rangle ,$

where $$\mathbf{K} = \left\langle {{k_1},{k_2}} \right\rangle$$ is a constant vector depending on the initial conditions. We determine this vector from the initial condition $$\mathbf{v}\left( 0 \right) = \left\langle {2,2} \right\rangle :$$

$\mathbf{v}\left( 0 \right) = \left\langle {3{t^2} + {k_1},4t + {k_2}} \right\rangle = \left\langle {0 + {k_1},0 + {k_2}} \right\rangle = \left\langle {{k_1},{k_2}} \right\rangle = \left\langle {2,2} \right\rangle .$

Hence, the particle's velocity is defined by the vector

$\mathbf{v}\left( t \right) = \left\langle {3{t^2} + 2,4t + 2} \right\rangle .$

Now we integrate the velocity vector to get the position vector:

$\mathbf{r}\left( t \right) = \int {\mathbf{v}\left( t \right)dt} = \int {\left\langle {3{t^2} + 2,4t + 2} \right\rangle dt} = \left\langle {{t^3} + 2t + {m_1},\,2{t^2} + 2t + {m_2}} \right\rangle .$

The constant vector $$\mathbf{M} = \left\langle {{m_1},{m_2}} \right\rangle$$ also depends on the initial condition. As $$\mathbf{r}\left( 0 \right) = \left\langle {0,0} \right\rangle ,$$ we have $$\mathbf{M} = \left\langle {{0},{0}} \right\rangle .$$

So the position vector is given by

$\mathbf{r}\left( t \right) = \left\langle {{t^3} + 2t,2{t^2} + 2t} \right\rangle .$

Calculate the coordinates of the particle at $$t = 1:$$

$\mathbf{r}\left( {t = 1} \right) = \left\langle {{1^3} + 2 \cdot 1,2 \cdot {1^2} + 2 \cdot 1} \right\rangle = \left\langle {3,4} \right\rangle .$

Then the displacement of the particle is equal

$\left| {\mathbf{r}\left( {t = 1} \right)} \right| = \sqrt {{3^2} + {4^2}} = 5\,\text{m}.$