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Q. 21

Expert-verified
Found in: Page 362

Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861

Use integration formulas to solve each integral in Exercises 21–62. You may have to use algebra, educated guess and-check, and/or recognize an integrand as the result of a product, quotient, or chain rule calculation. Check each of your answers by differentiating. $\int \left({x}^{2}-3{x}^{5}-7\right)dx$

The answer is $\frac{1}{3}{x}^{3}-\left(\frac{1}{2}\right){x}^{6}-7x+C.$

See the step by step solution

Step 1. Given information.

The given integral is $\int \left({x}^{2}-3{x}^{5}-7\right)dx.$

Step 2. Integration.

On integrating,

$\int \left({x}^{2}-3{x}^{5}-7\right)dx\phantom{\rule{0ex}{0ex}}=\int {x}^{2}dx-\int 3{x}^{5}dx-\int 7dx\phantom{\rule{0ex}{0ex}}=\frac{1}{3}{x}^{3}-3\left(\frac{1}{6}\right){x}^{6}-7x+C\phantom{\rule{0ex}{0ex}}=\frac{1}{3}{x}^{3}-\left(\frac{1}{2}\right){x}^{6}-7x+C$

Step 3. Verification.

On Differentiating$\frac{1}{3}{x}^{3}-\left(\frac{1}{2}\right){x}^{6}-7x+C$, we get,

$\frac{1}{3}×3{x}^{2}-\frac{1}{2}×6{x}^{5}-7\phantom{\rule{0ex}{0ex}}={x}^{2}-3{x}^{5}-7$

Hence proved.