Solved on Feb 09, 2024

Find the antiderivative of $\frac{d s}{d t}=11 t\left(5 t^{2}-7\right)^{3}$ .

STEP 1

Assumptions
1. We are given the derivative of a function $s$ with respect to $t$ , which is $\frac{d s}{d t}=11 t\left(5 t^{2}-7\right)^{3}$ .
2. We need to find the antiderivative of this function, which is the original function $s$ .

STEP 2

Recognize that the given derivative is in a form suitable for the reverse chain rule, also known as the substitution method for integration. The reverse chain rule is used when the integrand (the expression we are integrating) is the product of a function and the derivative of an inner function.

STEP 3

Identify the inner function and its derivative. In this case, the inner function is $u = 5t^2 - 7$ , and its derivative with respect to $t$ is $\frac{du}{dt} = 10t$ .

STEP 4

Notice that the given derivative $\frac{d s}{d t}$ contains the term $11t$ , which is similar to the derivative of the inner function $\frac{du}{dt} = 10t$ , except for the constant factor of $11$ instead of $10$ .

STEP 5

Rewrite the derivative $\frac{d s}{d t}$ in terms of $u$ by substituting $u = 5t^2 - 7$ and adjusting the constant factor to account for the difference between $11t$ and $10t$ .
$\frac{d s}{d t} = 11t(u)^3 = \frac{11}{10} \cdot 10t(u)^3$

STEP 6

Now, express $10t$ as $\frac{du}{dt}$ to make the substitution more explicit.
$\frac{d s}{d t} = \frac{11}{10} \cdot \frac{du}{dt}(u)^3$

STEP 7

Integrate both sides with respect to $t$ to find the antiderivative $s$ .
$\int \frac{d s}{d t} \, dt = \int \frac{11}{10} \cdot \frac{du}{dt}(u)^3 \, dt$

STEP 8

On the left side, the integral of $\frac{d s}{d t}$ with respect to $t$ is simply $s$ , since the derivative of $s$ with respect to $t$ is $\frac{d s}{d t}$ .
$s = \int \frac{11}{10} \cdot \frac{du}{dt}(u)^3 \, dt$

STEP 9

Perform the substitution $\frac{du}{dt} \, dt = du$ on the right side.
$s = \int \frac{11}{10} (u)^3 \, du$

STEP 10

Integrate $\frac{11}{10} (u)^3$ with respect to $u$ using the power rule for integration, which states that $\int u^n \, du = \frac{u^{n+1}}{n+1} + C$ , where $C$ is the constant of integration.
$s = \frac{11}{10} \int (u)^3 \, du$

STEP 11

Apply the power rule to the integral.
$s = \frac{11}{10} \cdot \frac{(u)^{3+1}}{3+1} + C$

STEP 12

Simplify the expression.
$s = \frac{11}{10} \cdot \frac{(u)^4}{4} + C$
$s = \frac{11}{40} (u)^4 + C$

STEP 13

Substitute back the expression for $u$ in terms of $t$ to get $s$ in terms of $t$ .
$s = \frac{11}{40} (5t^2 - 7)^4 + C$
This is the antiderivative of the given derivative.

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Find the antiderivative of $\frac{d s}{d t}=11 t\left(5 t^{2}-7\right)^{3}$ .

STEP 1

Assumptions
1. We are given the derivative of a function $s$ with respect to $t$ , which is $\frac{d s}{d t}=11 t\left(5 t^{2}-7\right)^{3}$ .
2. We need to find the antiderivative of this function, which is the original function $s$ .

STEP 2

Recognize that the given derivative is in a form suitable for the reverse chain rule, also known as the substitution method for integration. The reverse chain rule is used when the integrand (the expression we are integrating) is the product of a function and the derivative of an inner function.

STEP 3

Identify the inner function and its derivative. In this case, the inner function is $u = 5t^2 - 7$ , and its derivative with respect to $t$ is $\frac{du}{dt} = 10t$ .

STEP 4

Notice that the given derivative $\frac{d s}{d t}$ contains the term $11t$ , which is similar to the derivative of the inner function $\frac{du}{dt} = 10t$ , except for the constant factor of $11$ instead of $10$ .

STEP 5

Rewrite the derivative $\frac{d s}{d t}$ in terms of $u$ by substituting $u = 5t^2 - 7$ and adjusting the constant factor to account for the difference between $11t$ and $10t$ .
$\frac{d s}{d t} = 11t(u)^3 = \frac{11}{10} \cdot 10t(u)^3$

STEP 6

Now, express $10t$ as $\frac{du}{dt}$ to make the substitution more explicit.
$\frac{d s}{d t} = \frac{11}{10} \cdot \frac{du}{dt}(u)^3$

STEP 7

Integrate both sides with respect to $t$ to find the antiderivative $s$ .
$\int \frac{d s}{d t} \, dt = \int \frac{11}{10} \cdot \frac{du}{dt}(u)^3 \, dt$

STEP 8

On the left side, the integral of $\frac{d s}{d t}$ with respect to $t$ is simply $s$ , since the derivative of $s$ with respect to $t$ is $\frac{d s}{d t}$ .
$s = \int \frac{11}{10} \cdot \frac{du}{dt}(u)^3 \, dt$

STEP 9

Perform the substitution $\frac{du}{dt} \, dt = du$ on the right side.
$s = \int \frac{11}{10} (u)^3 \, du$

STEP 10

Integrate $\frac{11}{10} (u)^3$ with respect to $u$ using the power rule for integration, which states that $\int u^n \, du = \frac{u^{n+1}}{n+1} + C$ , where $C$ is the constant of integration.
$s = \frac{11}{10} \int (u)^3 \, du$

STEP 11

Apply the power rule to the integral.
$s = \frac{11}{10} \cdot \frac{(u)^{3+1}}{3+1} + C$

STEP 12

Simplify the expression.
$s = \frac{11}{10} \cdot \frac{(u)^4}{4} + C$
$s = \frac{11}{40} (u)^4 + C$

STEP 13

Substitute back the expression for $u$ in terms of $t$ to get $s$ in terms of $t$ .
$s = \frac{11}{40} (5t^2 - 7)^4 + C$
This is the antiderivative of the given derivative.

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Find the antiderivative of dsdt=11t(5t2−7)3\frac{d s}{d t}=11 t\left(5 t^{2}-7\right)^{3}dtds​=11t(5t2−7)3.

STEP 1

Assumptions1. We are given the derivative of a function s s s with respect to t t t, which is dsdt=11t(5t2−7)3 \frac{d s}{d t}=11 t\left(5 t^{2}-7\right)^{3} dtds​=11t(5t2−7)3.2. We need to find the antiderivative of this function, which is the original function s s s.

STEP 2

Recognize that the given derivative is in a form suitable for the reverse chain rule, also known as the substitution method for integration. The reverse chain rule is used when the integrand (the expression we are integrating) is the product of a function and the derivative of an inner function.

STEP 3

Identify the inner function and its derivative. In this case, the inner function is u=5t2−7 u = 5t^2 - 7 u=5t2−7, and its derivative with respect to t t t is dudt=10t \frac{du}{dt} = 10t dtdu​=10t.

STEP 4

Notice that the given derivative dsdt \frac{d s}{d t} dtds​ contains the term 11t 11t 11t, which is similar to the derivative of the inner function dudt=10t \frac{du}{dt} = 10t dtdu​=10t, except for the constant factor of 11 11 11 instead of 10 10 10.

STEP 5

STEP 6

Now, express 10t 10t 10t as dudt \frac{du}{dt} dtdu​ to make the substitution more explicit.dsdt=1110⋅dudt(u)3 \frac{d s}{d t} = \frac{11}{10} \cdot \frac{du}{dt}(u)^3 dtds​=1011​⋅dtdu​(u)3

STEP 7

Integrate both sides with respect to t t t to find the antiderivative s s s.∫dsdt dt=∫1110⋅dudt(u)3 dt \int \frac{d s}{d t} \, dt = \int \frac{11}{10} \cdot \frac{du}{dt}(u)^3 \, dt ∫dtds​dt=∫1011​⋅dtdu​(u)3dt

STEP 8

On the left side, the integral of dsdt \frac{d s}{d t} dtds​ with respect to t t t is simply s s s, since the derivative of s s s with respect to t t t is dsdt \frac{d s}{d t} dtds​.s=∫1110⋅dudt(u)3 dt s = \int \frac{11}{10} \cdot \frac{du}{dt}(u)^3 \, dt s=∫1011​⋅dtdu​(u)3dt

STEP 9

Perform the substitution dudt dt=du \frac{du}{dt} \, dt = du dtdu​dt=du on the right side.s=∫1110(u)3 du s = \int \frac{11}{10} (u)^3 \, du s=∫1011​(u)3du

STEP 10

STEP 11

Apply the power rule to the integral.s=1110⋅(u)3+13+1+C s = \frac{11}{10} \cdot \frac{(u)^{3+1}}{3+1} + C s=1011​⋅3+1(u)3+1​+C

STEP 12

Simplify the expression.s=1110⋅(u)44+C s = \frac{11}{10} \cdot \frac{(u)^4}{4} + C s=1011​⋅4(u)4​+Cs=1140(u)4+C s = \frac{11}{40} (u)^4 + C s=4011​(u)4+C

STEP 13

Substitute back the expression for u u u in terms of t t t to get s s s in terms of t t t.s=1140(5t2−7)4+C s = \frac{11}{40} (5t^2 - 7)^4 + C s=4011​(5t2−7)4+CThis is the antiderivative of the given derivative.

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Find the antiderivative of $\frac{d s}{d t}=11 t\left(5 t^{2}-7\right)^{3}$ .

Assumptions
1. We are given the derivative of a function $s$ with respect to $t$ , which is $\frac{d s}{d t}=11 t\left(5 t^{2}-7\right)^{3}$ .
2. We need to find the antiderivative of this function, which is the original function $s$ .

Identify the inner function and its derivative. In this case, the inner function is $u = 5t^2 - 7$ , and its derivative with respect to $t$ is $\frac{du}{dt} = 10t$ .

Notice that the given derivative $\frac{d s}{d t}$ contains the term $11t$ , which is similar to the derivative of the inner function $\frac{du}{dt} = 10t$ , except for the constant factor of $11$ instead of $10$ .

Now, express $10t$ as $\frac{du}{dt}$ to make the substitution more explicit.
$\frac{d s}{d t} = \frac{11}{10} \cdot \frac{du}{dt}(u)^3$

Integrate both sides with respect to $t$ to find the antiderivative $s$ .
$\int \frac{d s}{d t} \, dt = \int \frac{11}{10} \cdot \frac{du}{dt}(u)^3 \, dt$

On the left side, the integral of $\frac{d s}{d t}$ with respect to $t$ is simply $s$ , since the derivative of $s$ with respect to $t$ is $\frac{d s}{d t}$ .
$s = \int \frac{11}{10} \cdot \frac{du}{dt}(u)^3 \, dt$

Perform the substitution $\frac{du}{dt} \, dt = du$ on the right side.
$s = \int \frac{11}{10} (u)^3 \, du$

Apply the power rule to the integral.
$s = \frac{11}{10} \cdot \frac{(u)^{3+1}}{3+1} + C$

Simplify the expression.
$s = \frac{11}{10} \cdot \frac{(u)^4}{4} + C$
$s = \frac{11}{40} (u)^4 + C$

Substitute back the expression for $u$ in terms of $t$ to get $s$ in terms of $t$ .
$s = \frac{11}{40} (5t^2 - 7)^4 + C$
This is the antiderivative of the given derivative.