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Chapter 10: Q. 8TF (page 777)

Finding antiderivatives by undoing the chain rule: For each function f that follows, find a function F with the property that $F (x) = f (x)$ . You may have to guess and check to find such a function.
role="math" localid="1654760581753" $f (x) = x^{2} \sqrt{1 + x^{3}}$

Short Answer

Expert verified

The final integrate function is $\frac{1}{3} 鈭� \sqrt{t} dt = \frac{2}{9} {[1 + x^{3}]}^{\frac{3}{2}} + C$

Step by step solution

01

Given Information

The given function is $f (x) = x^{2} \sqrt{1 + x^{3}}$ .

02

Step 2. Integrate the given function.

The given function is $f (x) = x^{2} \sqrt{1 + x^{3}} . .. (i)$

$F' (x) = f (x)$

Adding up the numbers (1).

$鈭� x^{2} \sqrt{1 + x^{3}} dx$

Multiplying and dividing by 3.

$\frac{1}{3} 鈭� 3 x^{2} \sqrt{1 + x^{3}} dx . ... (ii)$

Assume that $1 + x^{3} = t$ and distinguish in terms of x.

$3 x^{3} dx = dt$

Replace the value in equation (2).

$\frac{1}{3} 鈭� \sqrt{t} dt$

Now it's time to integrate.

$\begin{array}{l} \frac{1}{3} 鈭� \sqrt{t} dt = \frac{1}{3} \frac{{[t]}^{\frac{1}{2} + 1}}{^{\frac{1}{2} + 1}} + C \\ \frac{1}{3} 鈭� \sqrt{t} dt = \frac{1}{3} 脳 (\frac{2}{3}) {[t]}^{\frac{3}{2}} + C \\ \frac{1}{3} 鈭� \sqrt{t} dt = \frac{2}{9} {[t]}^{\frac{3}{2}} + C \end{array}$

And,C denotes the integration constant.

Adding the value of t to the equation above.

$\frac{1}{3} 鈭� \sqrt{t} dt = \frac{2}{9} {[1 + x^{3}]}^{\frac{3}{2}} + C$

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Most popular questions from this chapter

Find $u + v a n d u - v$ also sketch $u, v, u + v, u - v$

role="math" localid="1649595165778" $u = (1, - 2) a n d v = (- 3, 6)$

If u, v and w are three vectors in ${鈩�}^{3}$ , which of the following products make sense and which do not?

localid="1649346164463" $(a) u 路 (v 路 w) (b) u 路 (v 脳 w) (c) u 脳 (v 路 w) (d) u 脳 (v 脳 w)$

If u and v are nonzero vectors in ${鈩�}^{3}$ , what is the geometric relationship between $u, v$ and $u 脳 v$ ?

In Exercises 30鈥�35 compute the indicated quantities when $u = (鈭� 3, 1, 鈭� 4), v = (2, 0, 5), and w = (1, 3, 13)$

Find the area of the parallelogram determined by the vectors u and v.

In Exercises 36鈥�41 use the given sets of points to find:

(a) A nonzero vector N perpendicular to the plane determined by the points.

(b) Two unit vectors perpendicular to the plane determined by the points.

(c) The area of the triangle determined by the points.

$P (2, 鈭� 5, 1), Q (鈭� 4, 5, 8), R (鈭� 1, 鈭� 5, 3)$

See all solutions

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