The factor x3 is grouped with 1+x4 because the derivative of the inner function 1+x4 is 4x3. This presence of x3 makes the expression x31+x4 directly integrable using substitution, which is a prerequisite for the integration by parts method.
Conditions: The integrand is x71+x4.; The method involves integration by parts.; The term dv must be integrable.
The factor x3 is grouped with 1+x4 because the derivative of the inner function 1+x4 is 4x3. This presence of x3 makes the expression x31+x4 directly integrable using substitution, which is a prerequisite for the integration by parts method.
Conditions: The integrand is x71+x4.; The method involves integration by parts.; The term dv must be integrable.
The substitution x=1+eu simplifies the integrand because its derivative dx=eudu exactly matches the eudu term in the original integral. This allows the complex expression eu1+eudu to be rewritten as the simpler radical xdx.
Conditions: The integrand contains eu1+eu.; The substitution is x=1+eu.; The differential is dx=eudu.
The substitution x=1+eu simplifies the integrand because its derivative dx=eudu exactly matches the eudu term in the original integral. This allows the complex expression eu1+eudu to be rewritten as the simpler radical xdx.
Conditions: The integrand contains eu1+eu.; The substitution is x=1+eu.; The differential is dx=eudu.
The antiderivative of x is 32x3/2. This is found by rewriting x as x1/2 and applying the power rule for integration, which states that ∫xndx=n+1xn+1 for n=−1.
Conditions: The function is x.; The power rule for integration is applied.; The exponent n=1/2.
The antiderivative of x is 32x3/2. This is found by rewriting x as x1/2 and applying the power rule for integration, which states that ∫xndx=n+1xn+1 for n=−1.
Conditions: The function is x.; The power rule for integration is applied.; The exponent n=1/2.
The antiderivative of u is found by rewriting the square root as a fractional power, u1/2, and then applying the power rule for integration. The power rule states that ∫undu=n+1un+1.
Conditions: The integrand is u, which is equivalent to u1/2.; The power rule for integration is applicable.; u≥0 for the real-valued square root form.
The antiderivative of u is found by rewriting the square root as a fractional power, u1/2, and then applying the power rule for integration. The power rule states that ∫undu=n+1un+1.
Conditions: The integrand is u, which is equivalent to u1/2.; The power rule for integration is applicable.; u≥0 for the real-valued square root form.
To set up the definite integral for the arc length of the curve y=x3/2 over the interval [0,32/9], you first apply the power rule to find the derivative of the function, which is f′(x)=23x1/2. Next, you square this derivative to get (f′(x))2=49x.
Conditions: The curve is defined by the function f(x)=x3/2.; The interval of integration is [0,32/9].; The arc length formula ∫ab1+(f′(x))2dx is applicable.
To set up the definite integral for the arc length of the curve y=x3/2 over the interval [0,32/9], you first apply the power rule to find the derivative of the function, which is f′(x)=23x1/2. Next, you square this derivative to get (f′(x))2=49x.
Conditions: The curve is defined by the function f(x)=x3/2.; The interval of integration is [0,32/9].; The arc length formula ∫ab1+(f′(x))2dx is applicable.
The derivative of the upper limit u=t with respect to t is 1/(2t). This factor is used in the chain rule when differentiating an integral with this variable upper limit.
Conditions: t>0.; The upper limit is defined as u=t.
The derivative of the upper limit u=t with respect to t is 1/(2t). This factor is used in the chain rule when differentiating an integral with this variable upper limit.
Conditions: t>0.; The upper limit is defined as u=t.
The substitution u=1+49x is chosen because it is exactly the expression under the radical in the arc-length integral ∫032/91+49xdx. By setting u to this inner expression, the complicated integrand simplifies to u, which is much easier to integrate using the power rule.
Conditions: The integral to evaluate is ∫032/91+49xdx.; The integrand contains a composite expression under a square root.
The substitution u=1+49x is chosen because it is exactly the expression under the radical in the arc-length integral ∫032/91+49xdx. By setting u to this inner expression, the complicated integrand simplifies to u, which is much easier to integrate using the power rule.
Conditions: The integral to evaluate is ∫032/91+49xdx.; The integrand contains a composite expression under a square root.